Question

The reaction taking place in a galvanic cell is as given
A(s) + B\textsuperscript{2+} (11100\textsuperscript{-3}M) $\rightarrow$ B(s) + A\textsuperscript{2+} (0.1M)
The emf of the cell is +2.651 V. If the standard emf of the cell is +2.71 V, what is the value of X?

• A. X = 6
• B. X = 2
• C. X = 3
• D. X = 4

Hint:

In galvanic cells, the Nernst equation allows us to calculate the cell potential at non-standard conditions by taking into account the concentrations of the ions involve(D)

Answer 1

The Correct Option is C

The equation for calculating the emf of a galvanic cell is given by the Nernst equation: \[ E = E^\circ - \dfrac{0.0591}{n} \log \dfrac{[A^{2+}]}{[B^{2+}]} \] Where: - \(E\) is the emf of the cell - \(E^\circ\) is the standard emf of the cell - \(n\) is the number of moles of electrons transferred (which we need to find) - \([A^{2+}]\) and \([B^{2+}]\) are the concentrations of \(A^{2+}\) and \(B^{2+}\) ions Given: - \(E = 2.651\) V (emf of the cell) - \(E^\circ = 2.71\) V (standard emf of the cell) - \([A^{2+}] = 0.1\) M - \([B^{2+}] = 11100\textsuperscript{-3}\) M (which simplifies to \(0.00111\) M) We can substitute these values into the Nernst equation and solve for \(n\): \[ 2.651 = 2.71 - \dfrac{0.0591}{n} \log \dfrac{0.1}{0.00111} \] First, calculate the log term: \[ \log \dfrac{0.1}{0.00111} = \log 90.09 \approx 1.954 \] Now substitute this into the equation: \[ 2.651 = 2.71 - \dfrac{0.0591}{n} \times 1.954 \] Solve for \(n\): \[ 2.71 - 2.651 = \dfrac{0.0591 \times 1.954}{n} \] \[ 0.059 = \dfrac{0.1156}{n} \] \[ n = \dfrac{0.1156}{0.059} \approx 3 \] Thus, the value of \(X\) is 3.