The magnitude of the emf of the following cell \(at ~25^{\circ} \mathrm{C} \) is:
\(P t\left(H_2\right) \mid 0.1 M\text{ acetic acid+ 0.1 M sodium acetate} ~\\\| ~0.1 \mathrm{M}\text{ formic acid + 0.1 M}\text{ sodium formate} \mid P t\left(H_2\right) \)

\( \left(\text { Use } K_{a(\text { acetic acid })}=1.8 \times 10^{-5} \text { and } K_{a(\text { Formic acid })} \\=2.1 \times 10^{-4}\right)\) 
1. 0.0315 volt
2. 0.0629 volt
3. 0.0455 volt
4. 0.0545 volt

Following cell has EMF 0.7995 V 

$Pt | H_2 (1 atm) | HN{O}_3 \left(1M\right) \left|\right| AgN{O}_3 \left(1M\right) | Ag$. 

If we add enough KCl to the Ag cell so that the final $C{l}^{-}$is 1M. Now the measured emf of the cell is 0.222 V. The $K_{SP}$ of AgCl would be :

$\left(1\right) 1 \times {10}^{–9.8 }$

$\left(2\right) 1 \times {10}^{–19.6}$

$\left(3\right) 2 \times {10}^{–10}$

$\left(4\right) 2.64 \times {10}^{–14}$

A flashlight cell has the cathodic reaction $2Mn{O}_2 \left(s\right) + Z{n}^{+2}+ 2\text{'}{e}^{-}\to Zn M{n}_2{O}_4 \left(S\right)$

If the flashlight cell is to give out 4.825 mA, how long could it run if initially 8.7 g of the

limiting reagent $Mn{O}_2$ is present? [Mn = 55, O = 16]

(1) $2 \times {10}^{6}sec$                                         

(2) $4 \times {10}^{6}sec$

(3) $6\times {10}^{6}sec$                                          

(4) $8 \times {10}^{6}sec$

An excess of granular zinc was added to 500 mL of 1M nickel nitrate till equilibrium was established. Find out the concentration of nickel at the equilibrium, if the standard electrode potential of $Z{n}^{2+} / Zn and N{i}^{2+}/Ni$ are – 0.75 and – 0.24 V, respectively.
(1) 5.56 x 10^-18 M
(2) 0.56 x 10^-18 M
(3) 2.16 x 10^-18 M
(4) 4.12 x 10^-18 M

Calculate the cell EMF in mV for $Pt | H_2 (1atm) | HCl (0.01 M) | AgCl(s) | Ag\left(s\right) at 298 K$ 

If $∆{G_r}^{°}$ values are at $25°C$ $-109.56\frac{kJ}{mol} for AgCl\left(s\right) \& –130.79\frac{kJ}{mol} for (H^{+}+ C{l}^{-}) \left(aq\right)$

(1) 456 mV                         

(2) 654 mV

(3) 546 mV                         

(4) None of these

A current of 0.1A was passed for 2hr through a solution of cuprocyanide and 0.3745 g of

copper was deposited on the cathode. Calculate the current efficiency for the copper

deposition. (Cu – 63.5)

1. 79%                                 

2. 39.5%

3. 63.25%                             

4. 63.5%

In acidic medium $Mn{O_4}^{-}$ is an oxidising agent. $Mn{O_4}^{-} + 8H^{+} + 5e^{-} \to M{n}^{2+} + 4H_{2}O. If H^{+}$ ion concentration is doubled, electrode potential of the half cell will : 

(1) Increase by 28.46 mV                             

(2) Decrease by 28.46 mV

(3) Increase by 14.23 mV                             

(4) Decrease by 142.30 mV

Calculate the EMF of the cell at 298 K $Pt | H_2 (1atm) | NaOH \left(xM\right), NaCl \left(xM\right) | AgCl (s) | Ag$ ${E^{°}}_{C{l}^{-}/AgCl/Ag}=+0.222 V$

(1) – 0.604 V                                                  

(2) – 0.04 V

(3) 1.048 V                                  

(4) emf depends on x and cannot be determined unless value of x is given

Assertion : Fluorine cannot be prepared from fluorides by chemical oxidation.

Reason : Fluorine is the strongest oxidizing agent due to its highly positive standard potential.

Assertion : : Sodium reacts with water at room temperature, whereas magnesium decomposes only hot water.

Reason : Sodium is more electropositive than magnesium.