Consider the following relations for emf of an electrochemical cell:

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

Which of the following combinations correctly represents the relation for the emf of the cell?

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

The correct expression that  represents the equivalent conductance at infinite dilution of $A{l}_2{\left(S{O}_4\right)}_3$ is:

(Given that ${\wedge }_{A{l}^{3+}}^{°}$ $and$ ${\wedge }_{S{O}_4^{2-}}^{°}$ are the equivalent conductances at infinite dilution of the respective ions)

1. ${\wedge }_{A{l}^{3+}}^{°}$ $+$ ${\wedge }_{S{O}_4^{2-}}^{°}$

2. $\left({\wedge }_{A{l}^{3+}}^{°}\right)$ $+$ ${\wedge }_{S{O}_4^{2-}}^{°}\times 6$

3. $\frac{1}{3}{\wedge }_{A{l}^{3+}}^{°}$ $+\frac{1}{2}$ ${\wedge }_{S{O}_4^{2-}}^{°}$

4. $2{\wedge }_{A{l}^{3+}}^{°}$ $+3$ ${\wedge }_{S{O}_4^{2-}}^{°}$

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 Gibb's energy for the decomposition of \(\mathrm{A l_{2} O_{3}}\) at \(\mathrm{500~ ^\circ C}\) is as follows: 

2/3Al₂O₃ → 4/3Al + O₂ ; ∆_rG = + 960 k J mol^–1

The potential difference needed for the electrolytic reduction of aluminium oxide (Al₂O₃) at \(\mathrm{500~ ^\circ C}\) is at least,

1. 3.0 V 

2. 2.5 V 

3. 5.0 V 

4. 4.5 V 

A hypothetical electrochemical cell is shown below.
A|A⁺(x M) || B⁺(y M)|B
The Emf measured is +0.20 V. The cell reaction is:

1. A⁺ + B → A + B⁺

2.  A⁺ + e^- → A ; B⁺ + e^- → B

3. The cell reaction cannot be predicted.

4. A + B⁺ → A⁺ + B

If ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\mathrm{o}}$ = -0.441 V and  ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^{\mathrm{o}}$ = 0.771 V, the standard emf of the reaction: 

Fe + 2Fe³⁺→ 3Fe²⁺ will be:

The efficiency of a fuel cell is given by:

1. $\frac{∆H}{∆G}$

2. $\frac{∆G}{∆S}$

3. $\frac{∆G}{∆H}$

4. $\frac{∆S}{∆G}$