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COMEDK · Chemistry · 18. Electrochemistry

For a cell \(2 \mathrm{M}_{(\mathrm{S})}+\mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}\) the \(\mathrm{E}^0\) cell \(=+1.67 \mathrm{~V}\). When \(\left[\mathrm{M}^{2+}\right]\) is \(1.0 \times 10^{-3} \mathrm{M}\) and \(\mathrm{p}\left(\mathrm{O}_2\right)\) is 0.1 atm , the EMF of the cell becomes +1.57 V . Calculate the pH of the electrochemical cell.

  1. A \(3.49\)
  2. B \(12.01\)
  3. C \(5.24\)
  4. D \(2.94\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2.94\)

Step-by-step Solution

Detailed explanation

The cell reaction is \(2M_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2M^{2+}_{(aq)} + 2H_2O_{(l)}\). The Nernst equation for the cell is \(E_{cell} = E^0_{cell} - \dfrac{0.0591}{n} \log Q\), where \(n=4\) electrons are transferred. The reaction quotient \(Q\) is given by…