The cell potential for the following cell notation is approximately
$M_{(s)} | M^{3+}(aq, 0.01 \ M) || N^{2+}(aq, 0.1 \ M) | N_{(s)}$
$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)

In the cell $Pt_{(s)} | H_2(g, 1 \, bar) | HCl_{(aq)} | AgCl_{(s)} | Ag_{(s)} | Pt_{(s)}$,the cell potential is $0.92 \, V$ when a $10^{-6} \, m$ $HCl$ solution is used. The standard electrode potential of the $(AgCl/Ag, Cl^-)$ electrode is ............. $V$ $\{ \text{Given, } \frac{2.303RT}{F} = 0.06 \, V \text{ at } 298 \, K \}$

The $emf$ of the following cell $Mg|Mg^{2+}(0.01 \ M)||Sn^{2+}(0.1 \ M)|Sn$ at $298 \ K$ in $V$ is: (Given: $E^{\circ}_{Mg^{2+}|Mg} = -2.34 \ V, E^{\circ}_{Sn^{2+}|Sn} = -0.14 \ V$)

For a Daniell cell,$E^0_{cell} = 1.1 \ V$. How is $K_c$ represented for the reaction occurring in the Daniell cell?

An electrochemical cell consists of the following two redox couples,$M^{x+}(aq)/M(s)$ $[E^{\ominus}_{red} = +0.15 \text{ V}]$ and $Fe^{3+}(aq)/Fe(s)$ $[E^{\ominus}_{red} = -0.036 \text{ V}]$. The cell $EMF$ is recorded to be $0.2057 \text{ V}$. If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$,then the value of $x$ is . . . . . . . (Nearest integer) [Given: $M$ is a $p$-block metal and $\frac{2.303RT}{F} = 0.059 \text{ V}$]

If $E^{\circ}(Cu^{2+}_{(aq)} \mid Cu_{(s)}) = +0.34 \ V$. What is the potential for $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} (0.1 \ M) + 2e^-$ at $298 \ K$?