Given the standard electrode potentials:

K+/K = –2.93 V

Ag+/Ag = 0.80 V

Hg2+/Hg = 0.79 V

Mg2+/Mg = –2.37 V

Cr3+/Cr = – 0.74 V

The correct increasing order of reducing power of the metals is:  

1.	Cr < Mg < K < Ag < Hg	2.	Mg < K < Ag < Hg < Cr
3.	K < Ag < Hg < Cr < Mg	4.	Ag < Hg < Cr < Mg < K

The correct statement about the given galvanic cell equation is -

Zn_(s) + 2Ag⁺­­­_(aq) → Zn²⁺_(aq) + 2Ag_(s)

 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-

The incorrect statement(s) among the below is/are:

1. Only (a)

2. (a) and (d)

3. (b) and (d)

4. (a) and (c)

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^–1. The molar conductivity will be -

The electricity required in coulombs for the oxidation of 1 mole of FeO to ${\mathrm{Fe}}_2{\mathrm{O}}_{\mathrm{3 }}$is:

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

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. –201 kJ

4. –31.12 kJ

The mass of nickel deposited by electrolysis of a solution of $\mathrm{Ni}{\left({\mathrm{NO}}_3\right)}_2$ using a current of 5 amperes for 20 min is: 
(Atomic mass of Nickle is 58.7u)

1. 2.42 g

2. 1.82 g

3. 3.93 g

4. 6.42 g

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

${\mathrm{E}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^{\mathrm{o}}=-2.36<mspace\; width="1em"></mspace>\mathrm{V};<mspace\; width="1em"></mspace>{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\mathrm{o}}=0.34\mathrm{V}$

The value of ${\mathrm{E}}_{\mathrm{cell}<mspace\; width="1em"></mspace>}$ for the above reaction is -