Q.57. One half cell in a voltaic cell is constructed by dipping silver rod in solution of unknown c

Q: What is the correct answer for JEE Main Chemistry Q57?

The correct answer is B. The cell consists of a zinc electrode and a silver electrode. Given standard electrode potentials: $E^\circ_{Zn^{2+}/Zn} = -0.76$ V (oxidation potential $E^\circ_{Zn/Zn^{2+}} = +0.76$ V) $E^\circ_{Ag^+/Ag} = +0.80$ V Since $E^\circ_{Ag^+/Ag}$ is higher than $E^\circ_{Zn^{2+}/Zn}$, silver will act as the cathode (reduction) and zinc will act as the anode (oxidation). Anode (Oxidation): $Zn(s) \ri

Q: Q.57. One half cell in a voltaic cell is constructed by dipping silver rod in solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of . A voltage of 1.60 V is…

The correct answer is B. The cell consists of a zinc electrode and a silver electrode. Given standard electrode potentials: $E^\circ_{Zn^{2+}/Zn} = -0.76$ V (oxidation potential $E^\circ_{Zn/Zn^{2+}} = +0.76$ V) $E^\circ_{Ag^+/Ag} = +0.80$ V Since $E^\circ_{Ag^+/Ag}$ is higher than $E^\circ_{Zn^{2+}/Zn}$, silver will act as the cathode (reduction) and zinc will act as the anode (oxidation). Anode (Oxidation): $Zn(s) \ri

Q: What are the options for this JEE Main Chemistry question?

A. $2/3.9$ · B. $4/5.9$ · C. $2.9/2$ · D. $5.9/4$

Question

Q.57. One half cell in a voltaic cell is constructed by dipping silver rod in  solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of . A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of Ag ions used in terms of log x (x = ) ? , ,  V

FirstInTest · Accepted Answer

The cell consists of a zinc electrode and a silver electrode.
Given standard electrode potentials:
$E^\circ_{Zn^{2+}/Zn} = -0.76$ V (oxidation potential $E^\circ_{Zn/Zn^{2+}} = +0.76$ V)
$E^\circ_{Ag^+/Ag} = +0.80$ V

Since $E^\circ_{Ag^+/Ag}$ is higher than $E^\circ_{Zn^{2+}/Zn}$, silver will act as the cathode (reduction) and zinc will act as the anode (oxidation).

Anode (Oxidation): $Zn(s) ightarrow Zn^{2+}(aq) + 2e^-$ ($E^\circ_{ox} = +0.76$ V)
Cathode (Reduction): $2Ag^+(aq) + 2e^- ightarrow 2Ag(s)$ ($E^\circ_{red} = +0.80$ V)
Overall cell reaction: $Zn(s) + 2Ag^+(aq) ightarrow Zn^{2+}(aq) + 2Ag(s)$

Standard cell potential $E^\circ_{cell} = E^\circ_{ox} + E^\circ_{red} = 0.76 	ext{ V} + 0.80 	ext{ V} = 1.56 	ext{ V}$.

Number of electrons transferred $n = 2$.

Using the Nernst equation:
$E_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log Q$
Given $E_{cell} = 1.60$ V and $\frac{2.303RT}{F} = 0.059$ V.
$Q = \frac{[Zn^{2+}]}{[Ag^+]^2}$

$1.60 = 1.56 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Ag^+]^2}$
Given $[Zn^{2+}] = 1$ M.
$1.60 = 1.56 - \frac{0.059}{2} \log \frac{1}{[Ag^+]^2}$
$1.60 - 1.56 = -\frac{0.059}{2} \log ([Ag^+]^{-2})$
$0.04 = -\frac{0.059}{2} (-2 \log [Ag^+])$ (using $\log a^b = b \log a$)
$0.04 = 0.059 \log [Ag^+]$

$\log [Ag^+] = \frac{0.04}{0.059} \approx 0.6779$.

The question asks for the concentration of Ag ions in terms of log x (where x = $[Ag^+]$). So, $\log [Ag^+]$ is the value we need. The closest option to 0.6779 is not immediately obvious. Let's recheck the options and calculation.

The options are given as fractions. Let's check the solution's calculation: $0.04 = +0.059 \log [Ag^+]$ (this is correct). Then it states $\log [Ag^+] = \frac{4}{5.9}$.
$\frac{4}{5.9} = \frac{40}{59} \approx 0.6779$. This matches our calculation.
So, the answer is $\frac{4}{5.9}$. This corresponds to option (2).

Answer

$2/3.9$

Answer

$4/5.9$

Answer

$2.9/2$

Answer

$5.9/4$

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