Electrochemistry MCQs for NEET — Chemistry Questions with Answers

(A) $\frac{8.7}{87}\times 96500=4.825\times {10}^{-3}{10}^{-3}\times t\Rightarrow t=2\times {10}^{6}sec$

(A) 

$$\begin{gathered} ∆G_{cell reaction}^{°}=2\left(-130.79\right)-2\left(-109.56\right) \\ =42.46 kJ/mole \\ \left(for H_2 + 2AgCl \to 2Ag + 2H^{+} + 2C{l}^{-}\right) \\ \therefore E{°}_{cell}=\frac{-42460}{-2\times 96500}=+0.220 V \\ Now, E_{cell}=+0.220+\frac{0.059}{2}\log \frac{1}{{\left(0.01\right)}^4} \\ =0.456 V=456mV \end{gathered}$$

(A) $$\begin{gathered} m \left(theoretical\right)=\frac{63.5\times 0.1\times 7200}{96500}=0.4738g \\ \therefore \% efficiency =\frac{0.3745}{0.4738}\times 100=79\% \end{gathered}$$

(A) $$\begin{gathered} E=E^{°}+\frac{0.0591}{5}\log {\left(2\right)}^8 \\ or E-E^{°}=\frac{0.0591}{5}\times 8 \log 2 = 0.02846 V = 28.46 mV \end{gathered}$$

(A) $$\begin{gathered} E_{B{r}^{-}/AgBr/Ag}^0=E_{A{g}^{+}/Ag}^0+\frac{0.059}{1}\log K_{SP} \\ AgBr=E_{A{g}^{+}/Ag}^0=-0.7257 \\ and E_{C{l}^{-}/AgCl/Ag}^0=E_{A{g}^{+}/Ag}^0+\frac{0.059}{1}\log K_{SP} \\ AgCl=E_{A{g}^{+}/Ag}^0=-0.59 \\ Now cell reaction is \\ Ag+B{r}^{-}\to AgBr+l{e}^{-} \\ AgCl +l{e}^{-}\to Ag+C{l}^{-} \\ B{r}^{-}+AgCl\stackrel{l{e}^{-}}{\to }C{l}^{-}+AgBr \\ 0=\left(0.7257-0.59\right)+\frac{0.059}{1}\log \frac{\left[B{r}^{-}\right]}{\left[C{l}^{-}\right]}\Rightarrow \frac{\left[B{r}^{-}\right]}{\left[C{l}^{-}\right]}=0.005 \end{gathered}$$

(A) Anode : $½ H_2 \to H^{+} + 1e^{-}$ 

Aathode : $AgCl + 1e^{-}\to Ag + C{l}^{-}$

$$\begin{gathered} \overline{Net: \frac{1}{2}{H}_2+AgCl\stackrel{1e^{-}}{\to }{H}^{+}+ Ag + C{l}^{-}} \\ {E}_{cell}=+0.222+\frac{0.059}{1}\log \frac{1}{\left[H^{+}\right]\left[C{l}^{-}\right]} \\ =+0.222+0.059 \log \frac{\left[O{H}^{+}\right]}{\left[{10}^{-14}\right]\left[C{l}^{-}\right]} \\ =+0.222+0.059\left(14\right)=+1.048volt \end{gathered}$$

$\left(C\right). Q = 10 \times 4825 = 48250 C$

no. of mole$=\frac{48250}{96500}=0.5$

$$\begin{gathered} Ag+\frac{1}{2}C{u}^{++}\to A{g}^{+}+\frac{1}{2}Cu \\ 2.00 2.00 \\ {E}_{cell}=E{°}_{cell}-\frac{0.0591}{1}\log \frac{\left[A{g}^{+}\right]}{{\left[C{u}^{++}\right]}^{1/2}} \\ {E}_1=E{°}_{cell}-\frac{0.0591}{1}\log \frac{2.00}{{\left(2.00\right)}^{1/2}} \\ {E}_2=E{°}_{cell}-\frac{0.0591}{1}\log \frac{2.00}{{\left(1.75\right)}^{1/2}} \\ ∆E=E_2-E_1=\frac{0.0591}{1}\left[\log \sqrt{2}-\log \frac{2.50}{\sqrt{1.75}}\right] \\ =\frac{0.0591}{1}\left[\log 1.41-\log 1.88\right]=\frac{0.0591}{1}\left[\log 1.492-0.2742\right] \\ =-\frac{0.0591}{1}\times 0.125=-0.00738 V \end{gathered}$$

A). Fluorine is the strongest oxidizing agent due to its highly positive standard potential. Therefore, fluorine has the highest tendency to get reduced to $F^{-}$, As a result, $F^{-}$ ion has the least tendency to get oxidised. That is why, fluorine cannot be prepared from fluorides by chemical oxidation.

Corrosion involves the surface of metals being coated with oxides or other salts due to oxidation. Rusting of iron, tarnishing of silver, and the green coating on copper (patina) are all examples of corrosion. Electrorefining of copper is an industrial process used to purify copper, where impure copper is oxidized at the anode and pure copper is deposited at the cathode; it is not a degradation process like corrosion. (Refer to the section 'Corrosion' and 'Electrorefining' in the context).

The context states, 'Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in the presence of H+...'. Therefore, H+ ions are essential for the cathodic reaction where oxygen is reduced. Oxidation of iron occurs at the anodic spot, and oxygen is reduced at the cathodic spot. Rust ($\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}$) is formed by the further oxidation of $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ by atmospheric oxygen. (Refer to the detailed explanation of iron corrosion chemistry).