For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]

  • A
    $4$
  • B
    $14$
  • C
    $28$
  • D
    $36$

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Similar Questions

The Nernst equation is given by $E = E^o - \frac{RT}{nF} \ln Q$. If $Q = K_{eq}$,then which of the following is not correct?

For the cell $Zn | Zn^{2+}_{(aq)} || Cu^{2+}_{(aq)} | Cu$,the standard cell potential $E^o$ is $1.10 \ V$ at $25^o \ C$. What is the order of magnitude of the equilibrium constant $K$ for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$?

Calculate the emf of the half-cell given below: $Pt(s) | H_2(g, 2 \text{ atm}) | HCl(aq, 0.02 \text{ M})$, $E^\circ_{H^+/H_2} = 0 \text{ V}$. (Given: $\frac{2.303RT}{F} = 0.059$, $\log 2 = 0.3010$)

The potential for the given half cell at $298 \ K$ is $(-) \ldots \ldots \ldots \times 10^{-2} \ V.$
$2 H^{+}_{(aq)} + 2 e^- \rightarrow H_{2(g)}$
$[H^{+}] = 1 \ M, P_{H_2} = 2 \ atm$
(Given: $2.303 RT / F = 0.06 \ V, \log 2 = 0.3$)

Consider the electrochemical cell: $Pt \ | \ O_{2(g)} \ (1 \ bar) \ | \ HCl \ (aq) \ || \ M^{2+} \ (aq, 1.0 \ M) \ | \ M_{(s)}$. The pH above which, oxygen gas would start to evolve at the anode is . . . . . . (nearest integer). $\left[ \text{Given :} \ E^{\circ}_{M^{2+}/M} = 0.994 \ V, \ E^{\circ}_{O_{2}/H_{2}O} = 1.23 \ V, \ \frac{RT}{F}(2.303) = 0.059 \ V \ \text{at the given condition} \right]$

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