Consider the following relations for emf of an electrochemical cell:

(a)	emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)
(b)	emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(c)	emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)
(d)	emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)

The correct relation among the given options is: 

1.	(a) and (b)	2.	(c) and (d)
3.	(b) and (d)	4.	(c) and (a)

Molar conductivities $(\wedge {°}_m)$ at infinite dilution of
NaCl, HCl, and $C{H}_{3}COONa$ are 126.4, 425.9, and 91.0 S cm² mol^-1 respectively.
 $(\wedge {°}_m)$  for $C{H}_{3}COOH$  will be: 

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

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. 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:

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

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

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

The value of E_cell in the reaction below will be:

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

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