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EMF of a cell from the Nernst equation at given concentrations

This type of question, asked 5 times: 2021 Paper 17, 2021 Paper 19, 2021 Paper 26, 30 Jan 2023, 22 Jan 2026 S1. All 5 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 17

    60
    For the cell Cu(s)∣Cu2+(aq)(0.1M)∥Ag+Cu(s) \mid Cu^{2 +}(aq)(0.1M) \parallel Ag^{+}(aq) (0.01M) \| Ag(s)Ag(s) the cell potential E1=0.3095 VE_{1}= 0.3095\text{ }V For the cell Cu(s)∣Cu2+Cu(s) \mid Cu^{2 +} (aq) (0.01 M) ∥Ag+\parallel Ag^{+}(aq) (0.001 M) ∣Ag(s)\mid Ag(s) the cell potential == ×10−2 V\times10^{- 2}\text{ }V. (Round off the Nearest Integer). [Use: 2.303RTF=0.059\frac{2.303RT}{F}= 0.059 ]

  2. Question 2

    2021 Paper 19

    55
    For the galvanic cell, Zn(s)+Cu2+(0.02 M)→Zn2+(0.04 M)+Cu(s)Zn(s) + Cu^{2 +}(0.02\text{ }M) \rightarrow Zn^{2 +}(0.04\text{ }M) + Cu(s), Ecell=____×10−2 VE_{\text{cell}}=\_\_\_\_\times10^{- 2}\text{ }V. (Nearest integer) [Use: ECu/Cu2+0=−0.34 VE_{Cu/Cu^{2 +}}^{0}= - 0.34\text{ }V, EZn/Zn2+0=+0.76 VE_{Zn/Zn^{2 +}}^{0}= + 0.76\text{ }V, 2.303RTF=0.059 V\frac{2.303RT}{F}= 0.059\text{ }V ]

  3. Question 3

    2021 Paper 26

    54
    Emf of the following cell at 298 K298\text{ }K in VV is x×10−2x \times 10^{- 2}.
    Zn∣Zn2+(0.1M)∣∣Ag+(0.01M)∣AgZn\left| Zn^{2 +}(0.1M) \right|\left| Ag^{+}(0.01M) \right|Ag
    The value of xx is ______. (Rounded off to the nearest integer) [Given: EZn2+/Zn0=−0.76 V;EAg+/Ag0=+0.80 V;2.303RT F=0.059E_{Zn^{2 +}/Zn}^{0}= - 0.76\text{ }V;E_{Ag^{+}/Ag}^{0}= + 0.80\text{ }V;\frac{2.303RT}{\text{ }F}= 0.059]

  4. Question 4

    30 Jan 2023

    146
    The electrode potential of the following half-cell at 298 K298\text{ }K X∣X2+(0.001M)∥Y2+(0.01M)∣YX\left| X^{2 +}(0.001M) \parallel Y^{2 +}(0.01M) \right|Y is ____×10−2 V\times10^{- 2}\text{ }V (Nearest integer). Given: EX2+∣X0=−2.36 VE_{X^{2 +}\mid X}^{0}= - 2.36\text{ }V
    EY2+∣Y0=+0.36 V2.303RTF=0.06 V\begin{matrix} & E_{Y^{2 +}\mid Y}^{0}= + 0.36\text{ }V \\ & \frac{2.303RT}{F}= 0.06\text{ }V \end{matrix}

  5. Question 5

    22 Jan 2026 S1

    47
    Consider the following electrochemical cell at 298 K Pt∣HSnO2−(aq)∣Sn(OH)6 2−(aq)∣Bi2O3(s)∣Bi(s)Pt\left| {HSnO}_{2}^{-}(aq) \right|Sn(OH)_{6}\ ^{2 -}(aq)\left| {Bi}_{2}O_{3}(s) \right|Bi(s). If the reaction quotient at a given time is 10610^{6}, then the cell EMF (Ecell E_{\text{cell~}}) is ____\_\_\_\_ ×10−1 V\times 10^{- 1}\text{ }V (Nearest integer). Given the standard half-cell reduction potential as EBi2O3/Bi,OH−0=−0.44 VE_{{Bi}_{2}O_{3}/{Bi}_{,{OH}^{-}}}^{0} = - 0.44\text{ }V and
    ESn(OH)62−/HSnO2−,OH0=−0.90 VE_{Sn(OH)_{6}^{2 -}/{HSnO}_{2}^{-},OH}^{0} = - 0.90\text{ }V