 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/4/b/iv/solution

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Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 4 / b / iv / Solution by

Codex  0 2026-10-20

At pH

3

,

[H^{+}] = 1 0^{- 3} M

, so the nitrate concentration is

0.001 + 2 (0.1) = 0.201 M

. With ideal thermodynamic activities,

Q = \frac{( 0.1 ) ^{3} ( 0.1 ) ^{2}}{( 0.201 ) ^{2} ( 1 0 ^{- 3} ) ^{8}} = 2.47519 \times 1 0^{20} .

Hence

E = 1.719 - \frac{8.3145 ( 298 )}{6 ( 96485 )} ln (2.47519 \times 1 0^{20}) = 1.5180 V .

The activity of nitric oxide is

0.1

because the supplied pressure is

0.1 bar

and the gas standard is

1 bar

. This numerical result additionally approximates zinc and nitrate thermodynamic activities by their standard-concentration ratios; without their activity coefficients, a more precise nonideal value is not specified.

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