For a $Mg|Mg^{2+}_{(aq)}||Ag^{+}_{(aq)}|Ag$ cell,the correct Nernst Equation is $:$

  • A
    $E_{\text{cell}} = E_{\text{cell}}^{o} - \frac{RT}{2F} \ln \frac{[Mg^{2+}]}{[Ag^{+}]^2}$
  • B
    $E_{\text{cell}} = E_{\text{cell}}^{o} + \frac{RT}{2F} \ln \frac{[Ag^{+}]^2}{[Mg^{2+}]}$
  • C
    $E_{\text{cell}} = E_{\text{cell}}^{o} - \frac{RT}{2F} \ln \frac{[Mg^{2+}]}{[Ag^{+}]}$
  • D
    $E_{\text{cell}} = E_{\text{cell}}^{o} - \frac{RT}{2F} \ln \frac{[Ag^{+}]^2}{[Mg^{2+}]}$

Explore More

Similar Questions

In acidic medium,$MnO_4^-$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$. If the $H^+$ ion concentration is doubled,the electrode potential of the half-cell will:

Derive an equation for the equilibrium constant $K_C$ of any galvanic cell (redox reaction) and also state its uses.

Difficult
View Solution

Consider the following electrodes $P = Zn^{2+}(0.0001 \ M) / Zn$,$Q = Zn^{2+}(0.1 \ M) / Zn$,$R = Zn^{2+}(0.01 \ M) / Zn$,$S = Zn^{2+}(0.001 \ M) / Zn$. Given $E^{\circ}(Zn^{2+} / Zn) = -0.76 \ V$,the electrode potentials of the above electrodes in volts are in the order:

The value of the reaction quotient $(Q)$ for the cell $Zn_{(s)} | Zn^{2+}(0.01 \ M) || Cu^{2+}(1.25 \ M) | Cu_{(s)}$ is:

For the cell at $298 \ K$:
$Ag_{(s)} | AgBr_{(s)} | Br^{-}(0.01 \ M) || I^{-}(0.02 \ M) | AgI_{(s)} | Ag_{(s)}$
The correct information is:
[Given: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo