The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?

For a cell involving a one-electron change at $25^o C$,$E^{o}_{cell} = 0.591 \ V$. The equilibrium constant for the reaction is .....

Calculate the equilibrium constant for the cell reaction at $298 \ K$: $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$,given $E^{0}_{cell} = 0.46 \ V$.

$E^0 = \frac{RT}{nF} \ln K_{eq}$. This is called

The cell potential for the following reaction is $0.03305 \ V$ at $298 \ K$. Find the value of $x$ for the reaction: $Zn | Zn^{2+} (0.1 \ M) || Cd^{2+} (x \ M) | Cd$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,$E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $M$)

$Cu^{2+} + 2e^- \to Cu$. On increasing $[Cu^{2+}]$ concentration,electrode potential