What is the oxidation potential of $0.05 \, M \, H_2SO_4$ in volts?

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

For a general redox reaction: $aA + bB \xrightarrow{n e^-} cC + dD$. Derive the Nernst equation.

What will be the $emf$ for the given cell $Pt|H_2(P_1)|H^{+}_{(aq)}||H_2(P_2)|Pt$?

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:

Calculate the cell potential for a $Cu$ plate kept in $0.2 \ M$ $CuSO_4$ solution. Given: $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$. (in $V$)