For the cell reaction with the indicated concentrations
$Al_{(s)} + 3Ag^{+}_{(aq)} \, (0.10 \, M) \to Al^{3+}_{(aq)} \, (0.30 \, M) + 3Ag_{(s)}$,
the measured voltage $(E_{cell})$ is $1.50 \, V$. Calculate $E^o_{cell} \, ........... \, V$
Given: $\frac{2.303 \, RT}{F} = 0.06 \, V ; \, \log \, 3 = 0.48$

  • A
    $1.5496$
  • B
    $1.654$
  • C
    $1.4032$
  • D
    None of these

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What will be the electrode potential of $Cu$ electrode dipped in $0.025 \ M$ $CuSO_4$ solution at $298 \ K$? Given that the standard reduction potential of $Cu^{2+}/Cu$ is $0.34 \ V$.

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

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: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]

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Calculate the electrode potential at $298 \; K$ for $Zn|Zn^{2+}$ electrode in which the activity of zinc ions is $0.001 \; M$ and $E^o_{Zn^{2+}/Zn}$ is $-0.76 \; V$. (in $; V$)

For the cell $Zn_{(s)} | Zn^{2+}_{(1\,M)} || Cu^{2+}_{(1\,M)} | Cu_{(s)}$,the $E^o_{cell} = 1.10 \, V$. When the cell is fully discharged at $298 \, K$,the ratio of the concentrations $[Zn^{2+}] / [Cu^{2+}]$ is:

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