What will be the electrode potential of a $Cu$ electrode dipped in $0.025 \ M \ CuSO_4$ solution at $298 \ K$,given that the standard reduction potential of $Cu$ is $0.34 \ V$ (in $V$)?

  • A
    $0.047$
  • B
    $0.293$
  • C
    $0.31$
  • D
    $0.387$

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Similar Questions

The reduction potential of a hydrogen electrode at $25^{\circ} C$ in a neutral solution is $(P_{H_2} = 1 \ atm)$. (in $V$)

What is the change in potential of the following cell $Zn_{(s)}|Zn^{2+} (1 \ M)||Pb^{2+} (1 \ M)|Pb_{(s)}$ if the concentration of ions at the anode is increased $10$ times?

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: $\frac{2.303 \ RT}{F} = 0.059$]

At $25\,^{\circ}C$,calculate the equilibrium constant for the cell reaction,
$X_{(s)} + Y_{(aq)}^{2+} \rightleftharpoons Y_{(s)} + X_{(aq)}^{2+}$
Given:
$E_{X^{2+}/X}^{o} = -1.36\,V$;
$E_{Y^{2+}/Y}^{o} = -0.76\,V$; $\frac{2.303\,RT}{F} = 0.06$

For a cell having a standard $emf$ of $0.295 \ V$ at $25^o \ C$ involving a two-electron change,the equilibrium constant for the reaction is:

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