For the given cell, Mg | Mg²⁺ || Cu²⁺ | Cu

a.  Mg acts as cathode

b.  Cu acts as cathode

c.  The cell reaction is   \(Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu\)

d.  Cu is the oxidising agent

The correct choice among the given is - 

The electrode potential for Mg electrode varies according to the equation

\(E_{Mg^{2+}/Mg}\ = \ E_{Mg^{2+}/Mg}^{o} \ - \ \frac{0.059}{2}log\frac{1}{[Mg^{2+}]}\) 

The graph of E_{Mg²⁺ / Mg} vs log [Mg²⁺] among the following is:

1.		2.
3.		4.

Find the emf of the cell in which the following reaction takes place at 298 K:
\(\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.001 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s}) \)
\( \small{\text { (Given that } \mathrm{E}_{\text {cell }}^{\circ}=10.5 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \text { at } \ 298 \mathrm{~K})} \)
1. 1.05 V
2. 1.0385 V 
3. 1.385 V
4. 0.9615 V 

The value of E_cell in the reaction below will be:

\(\small{Pt(s)|Br^{-}(0.010 \ M)|Br_{2}(l) \ ||H^{+}(0.030 \ M)|H_{2}(g)(1 \ bar)|Pt(s)}\)

\(E_{Br^{-}/Br_{2}}^{o} \ = \ -1.09 \ V\)
1. +1.298 V

Mg(s) + 2Ag⁺(0.0001M) → Mg²⁺(0.130M) + 2Ag(s)

If  E^Ɵ(cell) for the above mentioned cell is 3.17 V, then E(cell) value will be-

(log 13=1.1)

1. 2.87 V
2. 3.08 V
3. 2.96 V
4. 2.68 V

The emf of the cell in which the following reaction takes place is:

Ni_(s)+2Ag⁺(0.002 M) →Ni²⁺(0.160 M) + 2Ag_(s)
( Given that \(E_{cell}^{o}\)= 1.05 V)

1. E_cell= 9.14 V
2. E_cell= 0.0914 V
3. E_cell= 0.00914 V
4. E_cell= 0.914 V

${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$= 1.1 V for Daniel cell. Which of the following expressions are correct descriptions of the state of equilibrium in this cell?

(a) 1.1 = K_C

(b) $\frac{2.303\mathrm{RT}}{2\mathrm{F}}{\mathrm{logK}}_{\mathrm{C}}=1.1$

(c) ${\mathrm{logK}}_{\mathrm{C}}=\frac{2.2}{0.059}$

(d) log K_C = 1.1

The correct choice among the given is -

1. (a, b)

2. (b, c)

3. (c, d)

4. (a, d)

Mg_(s) | Mg²⁺_(0.001M) || Cu²⁺_{(0.0001 M)} | Cu_(s)

${\mathrm{E}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^{\mathrm{o}}=-2.36 \mathrm{V}; {\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\mathrm{o}}=0.34\mathrm{V}$

The value of ${\mathrm{E}}_{\mathrm{cell} }$ for the above reaction is -