\(Fe^{2+}(aq) \ + \ Ag^{+}(aq) \ \rightarrow \ Fe^{3+}(aq) \ + \ Ag(s)\)

\(E_{Fe^{3+}/Fe^{2+}}^{o} \ = \ 0.77 \ V; \ E_{Ag^{+}/Ag}^{o} \ = \ 0.80 \ V\)

The value of \(\Delta G_{r}^{o} \)  in the above reaction will be:

1. +$2.89$ $\mathrm{kJ}$

2. $-2.89$ $\mathrm{kJ}$

3. $-2.89$ $\mathrm{J}$

4. -186.83 J

Given:
(i) ${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$    E^o = 0.337 V 
(ii) ${\mathrm{Cu}}^{2+}+{\mathrm{e}}^{-}\to {\mathrm{Cu}}^{+}$  E^o = 0.153 V 
Electrode potential, E^o for the reaction, 
${\mathrm{Cu}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Cu}$, will be: 

1. 0.52 V

2. 0.90 V

3. 0.30 V

4. 0.38 V

For a reaction A(s) + 2B⁺ $\to$ A²⁺ + 2B(s) ;  K_C has been found to be 10¹². The ${\mathrm{E}}_{\mathrm{cell}}^{°}$ is : 

The value of  ∆G°  in the reaction below would be:
\(\small{\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_2 \mathrm{O}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Zn}^{+2}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq})}\)

Given: \(E_{cell}^{\circ} = 1.04V\)

1. –2.13 kJ

2. –21.3 kJ

3. –213.04 kJ

4. –31.12 kJ

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of the electrode would be: 

For the cell reaction \(2Fe^{3+}(aq) \ + \ 2I^{-}(aq)\rightarrow 2Fe^{2+}(aq) \ + \ I_{2}(aq)\)

\(E_{cell}^{o} \ = \ 0.24 \ V\) at $298$ $\mathrm{K}$. The standard Gibbs energy ∆_rG^⊝ of the cell reaction is:

[Given: $\mathrm{F }= 96500$ $\mathrm{C}$ ${\mathrm{mol}}^{-1}$]

1. $23.16$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

2. $-46.32$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

3. $-23.16$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

4. $46.32$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd

${{\mathrm{E}}^{⊝}}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}=$ $-0.74$ $\mathrm{V}$; ${{\mathrm{E}}^{⊝}}_{{\mathrm{Cd}}^{2+}/\mathrm{Cd}}=-0.40$ $\mathrm{V}$

The value of $∆{\mathrm{G}}_r^{\mathrm{o}}$ in the above reaction will be-

For a cell involving one electron ${\mathrm{E}}_{\mathrm{cell}}^{⊝}=0.\mathrm{59 }\mathrm{V}$ at 298 K.
The equilibrium constant for the cell reaction is :
( $\left[\mathrm{Given}\right]$ $\mathrm{that}$ $\frac{2.303}{}$ $\mathrm{RT}\mathrm{F}=0.059$ $\mathrm{V}$ $\mathrm{at}$ $\mathrm{T}=298$ $\mathrm{K)}$

1. $1.0\times {10}^{30}$

2. $1.0\times {10}^2$

3. $1.0\times {10}^5$

4. $1.0\times {10}^{10}$

Given cell: 

Pt(s)|H₂(g)(P = 1 atm)| CH₃COOH(0.1 M), HCl(0.1 M) || KCl(aq)|Hg₂Cl₂(s)|Hg

EMF of the cell is found to be 0.045 V at 298 K and the temperature coefficient is 3.4×10^-4 V K^-1. The cell entropy change of the following cell is : 

[Given K_a CH₃COOH = 10^-5  M]

1. 70.8 J K⁻¹ mol⁻¹                      

2. 65.2 J K⁻¹ mol⁻¹

3. 79.2 J K⁻¹ mol⁻¹                  

4. 83.5 J K⁻¹ mol⁻¹