What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

The $\text{EMF}$ of a galvanic cell consisting of two hydrogen electrodes is $0.17 \ V$. If the solution of one of the electrodes has $[H^+] = 10^{-3} \ M$,the $\text{pH}$ at the other electrode is:

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 $\frac{2.303RT}{F} = 0.059 \text{ V}$]

At $298 \ K$,find out the $emf$ for the cell:
$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$
Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

For a certain redox reaction in a galvanic cell $X(s) + Y^{2+}_{(aq)} \rightarrow X^{2+}_{(aq)} + Y(s)$, $E^0_{cell}$ is $0.0296 \text{ V}$ at $298 \text{ K}$. What is the equilibrium constant of the reaction?

Consider the cell:
$Pt_{(s)} | H_{2(g)}(1 \ atm) | H^{+}_{(aq)}, [H^{+}]=1 \ M || Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} | Pt_{(s)}$
Given: $E^0_{Fe^{3+}/Fe^{2+}} = 0.771 \ V$ and $E^0_{H^{+}/\frac{1}{2}H_2} = 0 \ V$ at $T = 298 \ K$.
If the potential of the cell is $0.712 \ V$,the ratio of concentration of $Fe^{2+}$ to $Fe^{3+}$ is $........$. (Nearest integer)