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NEET-UG Biology

Electrochemical cells, EMF and the electrochemical series — practice questions

93 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.

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  1. Question 1 · difficulty L2 · understanding

    The hydrogen electrode is dipped in a solution of pH=3\mathrm{pH}=3 at 25C25^{\circ} \mathrm{C}. The potential of the electrode will be _________ ×102 V\times 10^{-2} \mathrm{~V}. (2.303RTF=0.059 V)\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)

  2. Question 2 · difficulty L2 · understanding

    In a cell, the following reactions take place \begin{aligned}\matrix{ {F{e^{2 + }} \to F{e^{3 + }} + {e^ - }} & {E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.77\,V} \cr {2{I^ - } \to {I_2} + 2{e^ - }} & {E_{{I_2}/{I^ - }}^o = 0.54\,V} \cr } \end{aligned} The standard electrode potential for the spontaneous reaction in the cell is x ×\times 10 -2 V 298 K. The value of x is ____________. (Nearest Integer)

  3. Question 3 · difficulty L2 · understanding

    Given at 298 K : EFe2+/Fe=X Volt EFe3+/Fe=Y Volt  \begin{aligned} & \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\ominus}=\mathrm{X} \text { Volt } \\ & \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{\ominus}=\mathrm{Y} \text { Volt } \end{aligned} The EFe3+/Fe2+\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\ominus} in Volt at 298 K is given by :

    • A. 2X3Y2 \mathrm{X}-3 \mathrm{Y}
    • B. 3Y2X3 Y-2 X
    • C. 3Y+2X3 \mathrm{Y}+2 \mathrm{X}
    • D. Y+XY+X
  4. Question 4 · difficulty L2 · understanding

    Consider the following reduction processes : Al3++3eAl( s),E0=1.66 VFe3++eFe2+,E0=+0.77 VCo3++eCo2+,E0=+1.81 VCr3++3eCr( s),E0=0.74 V \begin{aligned} & \mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(\mathrm{~s}), \mathrm{E}^0=-1.66 \mathrm{~V} \\ & \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}, \mathrm{E}^0=+0.77 \mathrm{~V} \\ & \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+}, \mathrm{E}^0=+1.81 \mathrm{~V} \\ & \mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(\mathrm{~s}), \mathrm{E}^0=-0.74 \mathrm{~V} \end{aligned} The tendency to act as reducing agent decreases in the order :

    • A. Al>Fe2+>Cr>Co2+\mathrm{Al}>\mathrm{Fe}^{2+}>\mathrm{Cr}>\mathrm{Co}^{2+}
    • B. Al>Cr>Co2+>Fe2+\mathrm{Al}>\mathrm{Cr}>\mathrm{Co}^{2+}>\mathrm{Fe}^{2+}
    • C. Cr>Fe2+>Al>Co2+\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Al}>\mathrm{Co}^{2+}
    • D. Al>Cr>Fe2+>Co2+\mathrm{Al}>\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Co}^{2+}
  5. Question 5 · difficulty L2 · understanding

    The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.

    • A. ESn4+/Sn2+o=+1.15 VE^o_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +1.15 \text{ V}
    • B. EAl3+/Alo=1.66 VE^o_{\text{Al}^{3+}/\text{Al}} = -1.66 \text{ V}
    • C. EPb4+/Pb2+o=+1.67 VE^o_{\text{Pb}^{4+}/\text{Pb}^{2+}} = +1.67 \text{ V}
    • D. ETl3+/Tlo=+1.26 VE^o_{\text{Tl}^{3+}/\text{Tl}} = +1.26 \text{ V}
  6. Question 6 · difficulty L2 · understanding

    Based on the data given below : ECr2O72/Cr3+=1.33 VECl2/Cl()=1.36 VEMnO4/Mn2+0=1.51 VECr3+/Cr=0.74 V\begin{array}{ll} \mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\circ}=1.33 \mathrm{~V} & \mathrm{E}_{\mathrm{Cl}_2 / \mathrm{Cl}^{(-)}}^{\circ}=1.36 \mathrm{~V} \\ \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^0=1.51 \mathrm{~V} & \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V} \end{array} the strongest reducing agent is :

    • A. Cl\mathrm{Cl}^{-}
    • B. MnO4\mathrm{MnO}_4^{-}
    • C. Cr\mathrm{Cr}
    • D. Mn2+\mathrm{Mn}^{2+}
  7. Question 7 · difficulty L2 · understanding

    One of the commonly used electrode is calomel electrode. Under which of the following categories, calomel electrode comes?

    • A. Metal ion - Metal electrodes
    • B. Oxidation - Reduction electrodes
    • C. Metal - Insoluble Salt - Anion electrodes
    • D. Gas - Ion electrodes
  8. Question 8 · difficulty L2 · understanding

    Reduction potential of ions are given below: ClO4IO4BrO4E=1.19 VE=1.65 VE=1.74 V\begin{array}{ccc} \mathrm{ClO}_4^{-} & \mathrm{IO}_4^{-} & \mathrm{BrO}_4^{-} \\ \mathrm{E}^{\circ}=1.19 \mathrm{~V} & \mathrm{E}^{\circ}=1.65 \mathrm{~V} & \mathrm{E}^{\circ}=1.74 \mathrm{~V} \end{array} The correct order of their oxidising power is :

    • A. IO4>BrO4>ClO4\mathrm{IO}_4^{-}>\mathrm{BrO}_4^{-}>\mathrm{ClO}_4^{-}
    • B. BrO4>ClO4>IO4\mathrm{BrO}_4^{-}>\mathrm{ClO}_4^{-}>\mathrm{IO}_4^{-}
    • C. ClO4>IO4>BrO4\mathrm{ClO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{BrO}_4^{-}
    • D. BrO4>IO4>ClO4\mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}
  9. Question 9 · difficulty L2 · understanding

    In 3d series, the metal having the highest M 2+ /M standard electrode potential is :

    • A. Cr
    • B. Fe
    • C. Cu
    • D. Zn
  10. Question 10 · difficulty L2 · understanding

    The (ET)P{\left( {{{\partial E} \over {\partial T}}} \right)_P} of different types of half cells are as follows: A B C D 1×1041 \times {10^{ - 4}} 2×1042 \times {10^{ - 4}} 0.1×1040.1 \times {10^{ - 4}} 0.2×1040.2 \times {10^{ - 4}} (Where E is the electromotive force) Which of the above half cells would be preferred to be used as reference electrode?

    • A. A
    • B. B
    • C. C
    • D. D

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