Chemistry Q57 – JEE-MAIN 2026

Home Chemistry ElectrochemistryQuestion #389

One half cell in a voltaic cell is constructed by dipping silver rod in

A g N O_{3}

solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of

Z n S O_{4}

. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of

A g^{+}

ions used in terms of

\log x = [A g^{+}]

E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V

E_{A g^{+} / A g}^{\circ} = + 0.80 V

\frac{2.303 R T}{F} = 0.059 V

Clue 1 of 3: Cell Potential Calculation

Determine the standard cell potential ( $E_{c e l l}^{\circ}$ ) by identifying the anode and cathode and using their standard reduction potentials.

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● Video Walkthrough

Step 1: Determine Standard Cell Potential and Number of Electrons✦ Active

The standard reduction potentials are $E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V$ and $E_{A g^{+} / A g}^{\circ} = + 0.80 V$ . Zinc has a lower reduction potential, so it will be oxidized (anode), and silver ions will be reduced (cathode).

E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ} = E_{A g^{+} / A g}^{\circ} - E_{Z n^{2 +} / Z n}^{\circ} = (+ 0.80 V) - (- 0.76 V) = 1.56 V

The balanced overall cell reaction is $Z n (s) + 2 A g^{+} (a q) \to Z n^{2 +} (a q) + 2 A g (s)$ . From this, the number of electrons transferred, $n$ , is 2.

Step 2: Apply the Nernst Equation○ Expand

The Nernst equation at 298 K is $E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.059}{n} \log Q$ . We are given $E_{c e l l} = 1.60 V$ , and we calculated $E_{c e l l}^{\circ} = 1.56 V$ and $n = 2$ .

The reaction quotient $Q$ for the reaction is $Q = \frac{[Z n^{2 +}]}{[A g^{+}]^{2}}$ . We are given $[Z n^{2 +}] = 1 M$ and we need to find $[A g^{+}]$ . Let $[A g^{+}] = x$ . So, $Q = \frac{1}{x^{2}}$ .

1.60 = 1.56 - \frac{0.059}{2} \log (\frac{1}{x^{2}})

💡 Teacher's Secret Hint

Remember that the concentration of pure solids (Zn and Ag) are not included in the reaction quotient Q.

Step 3: Solve for

\log x

○ Expand

Rearrange the Nernst equation to solve for $\log x$ :

1.60 - 1.56 = - \frac{0.059}{2} \log (x^{- 2})

0.04 = - \frac{0.059}{2} (- 2 \log x)

0.04 = 0.059 \log x

\log x = \frac{0.04}{0.059} = \frac{40}{59}

Comparing this result with the given options, $\frac{40}{59}$ matches option 2.

💡 Teacher's Secret Hint

Pay attention to the sign changes when manipulating logarithmic terms, especially $\log (A^{- B}) = - B \log A$ .

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Chemistry Question 57 – JEE-MAIN 2026

Clue 1 of 3: Cell Potential Calculation

🧠 Full Solution Path