The standard $emf$ of a galvanic cell involving $3$ moles of electrons in its redox reaction is $0.59 \ V$. The equilibrium constant for the reaction of the cell is:

Which will increase the voltage of the cell $Sn_{(s)} + 2Ag^{+}_{(aq)} \to Sn^{2+}_{(aq)} + 2Ag_{(s)}$?

The equilibrium constant of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is approximately:

Calculate the cell potential at $298 \ K$ for the following cells:
$(a)$ $Cd \mid Cd^{2+}(0.02 \ M) \parallel H^{+}(1 \ M) \mid H_{2(g)}(1 \ bar) \mid Pt$ $\left[ E_{Cd^{2+} \mid Cd}^0 = -0.40 \ V \right]$
$(b)$ $Al \mid Al^{3+}(0.25 \ M) \parallel Zn^{2+}(0.15 \ M) \mid Zn_{(s)}$ $\left[ E_{Al^{3+} \mid Al}^0 = -1.66 \ V, E_{Zn^{2+} \mid Zn}^0 = -0.76 \ V \right]$

For the following electrochemical cell at $298 \ K$,
$Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 1 \ M) \parallel M^{4+}_{(aq)}, M^{2+}_{(aq)} \mid Pt_{(s)}$
$E_{\text{cell}} = 0.092 \ V$ when $\frac{[M^{2+}_{(aq)}]}{[M^{4+}_{(aq)}]} = 10^x$
Given : $E^0_{M^{4+}/M^{2+}} = 0.151 \ V$; $2.303 \frac{RT}{F} = 0.059 \ V$
The value of $x$ is

At what $pH$,given half cell $MnO_4^{-} (0.1 \ M) \mid Mn^{2+} (0.001 \ M)$ will have electrode potential of $1.282 \ V$? (Nearest Integer) Given $E_{MnO_4^{-} / Mn^{2+}}^{o} = 1.54 \ V, \frac{2.303 RT}{F} = 0.059 \ V$