The cell reaction of the given cell is spontaneous if:
$Pt | Cl_2 (P_1 \, atm) | Cl^{-} (1 \, M) || Cl^{-} (1 \, M) | Cl_2 (P_2 \, atm) | Pt$

Consider the cell whose $emf$ is $1.01 \ V$.
$Pt, H_2(1 \ atm) | H^{+}(pH = 4) || Ag^{+}(xM) | Ag$
What is the value of $x$? (Given: $E^o_{Ag^{+}|Ag} = +0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

For the redox reaction occurring in a cell: $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$,if $E^o_{cell} = 1.10 \ V$,calculate the value of $E_{cell}$ in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$)

The reduction potential of a hydrogen half-cell will be negative if:

The cell potential for the given cell at $298 \, K$ is $Pt \mid H_2 (g, 1 \, bar) \mid H^{+}_{(aq)} \parallel Cu^{2+}_{(aq)} \mid Cu_{(s)}$. The cell potential is $0.31 \, V$. The $pH$ of the acidic solution is $3$,and the concentration of $Cu^{2+}$ is $10^{-x} \, M$. The value of $x$ is $.....$ (Given: $E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \, V$ and $\frac{2.303 RT}{F} = 0.06 \, V$)

What will be the electromotive force of the following cell $Fe|Fe^{+2}\,(0.2\, M)\,||\, Au^{+3}\,(0.02\, M)\,|Au$ (in $,V$)?
Given: $E^o_{Fe^{+2}/Fe} = -0.44\,V$ and $E^o_{Au^{+3}/Au} = +1.50\,V$.