Chemistry Q75 – JEE-MAIN 2025

Home Chemistry ElectrochemistryQuestion #402

Consider the following half cell reaction

{Cr}_{2} O_{7}^{2 -} (aq) + 6 e^{-} + 14 H^{+} (aq) ⟶ 2 {Cr}^{3 +} (aq) + 7 H_{2} O (l)

The reaction was conducted with the ratio of

\frac{[{Cr}^{3 +}]^{2}}{[{Cr}_{2} O_{7}^{2 -}]} = 10^{- 6}

. The pH value at which the EMF of the half cell will become zero is _______. (nearest integer value) [Given : standard half cell reduction potential

E_{{Cr}_{2} O_{7}^{2 -}, H^{+} / {Cr}^{3 +}}^{\circ} = 1.33 V

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

Enter Calculated Numerical Value:

Clue 1 of 3: Nernst Equation

The Nernst equation relates the cell potential to the standard cell potential and the concentrations of reactants and products.

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Step 1: Apply the Nernst Equation✦ Active

The Nernst equation for the given half-cell reaction is $E = E^{\circ} - \frac{2.303 R T}{n F} \log Q$ . Given $E = 0 V$ , $E^{\circ} = 1.33 V$ , $n = 6$ , and $\frac{2.303 R T}{F} = 0.059 V$ . The reaction quotient $Q$ is given by $Q = \frac{[{Cr}^{3 +}]^{2}}{[{Cr}_{2} O_{7}^{2 -}] [H^{+}]^{14}}$ . Substituting these into the Nernst equation:

0 = 1.33 - \frac{0.059}{6} \log (\frac{[{Cr}^{3 +}]^{2}}{[{Cr}_{2} O_{7}^{2 -}] [H^{+}]^{14}})

💡 Teacher's Secret Hint

Ensure correct identification of 'n' (number of electrons) and the form of the reaction quotient 'Q'.

Step 2: Substitute given concentration ratio and simplify○ Expand

We are given $\frac{[{Cr}^{3 +}]^{2}}{[{Cr}_{2} O_{7}^{2 -}]} = 10^{- 6}$ . Substitute this into the equation:

0 = 1.33 - \frac{0.059}{6} \log (\frac{10^{- 6}}{[H^{+}]^{14}})

0 = 1.33 - \frac{0.059}{6} (\log (10^{- 6}) - \log ([H^{+}]^{14}))

0 = 1.33 - \frac{0.059}{6} (- 6 - 14 \log [H^{+}])

Using the definition

pH = - \log [H^{+}]

0 = 1.33 - \frac{0.059}{6} (- 6 + 14 pH)

💡 Teacher's Secret Hint

Remember the properties of logarithms, especially $\log (A / B) = \log A - \log B$ and $\log (A^{x}) = x \log A$ .

Step 3: Solve for pH○ Expand

Rearrange the equation to solve for pH:

1.33 = \frac{0.059}{6} (- 6 + 14 pH)

\frac{1.33 \times 6}{0.059} = - 6 + 14 pH

135.254 \approx - 6 + 14 pH

141.254 \approx 14 pH

pH = \frac{141.254}{14} \approx 10.089

Rounding to the nearest integer, the pH value is 10.

💡 Teacher's Secret Hint

Perform calculations carefully and round to the nearest integer as requested.

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Chemistry Question 75 – JEE-MAIN 2025

Clue 1 of 3: Nernst Equation

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