QuestionJuly 17, 2025

Calculate the cell potential, E_(cell) for the reaction below when [Cr^3+]=0.25M and [Pb^2+]=0.15M 2Cr(s)+3Pb^2+(aq)leftharpoons 2Cr^3+(aq)+3Pb(s) E_(cell)^circ =+0.61V E_(cell)=0.597V E_(cell)=0.573V E_(cell)=1.09V E_(cell)=1.21V E_(cell)=0.871V

Calculate the cell potential, E_(cell) for the reaction below when [Cr^3+]=0.25M and [Pb^2+]=0.15M 2Cr(s)+3Pb^2+(aq)leftharpoons 2Cr^3+(aq)+3Pb(s) E_(cell)^circ =+0.61V E_(cell)=0.597V E_(cell)=0.573V E_(cell)=1.09V E_(cell)=1.21V E_(cell)=0.871V
Calculate the cell potential, E_(cell) for the reaction below when [Cr^3+]=0.25M and [Pb^2+]=0.15M 2Cr(s)+3Pb^2+(aq)leftharpoons 2Cr^3+(aq)+3Pb(s) E_(cell)^circ =+0.61V E_(cell)=0.597V E_(cell)=0.573V E_(cell)=1.09V E_(cell)=1.21V E_(cell)=0.871V

Solution
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Answer

E_{cell} = 0.573V Explanation 1. Identify the Nernst Equation Use the Nernst equation: E_{cell} = E_{cell}^{\circ} - \frac{RT}{nF} \ln Q. For simplicity at room temperature, use E_{cell} = E_{cell}^{\circ} - \frac{0.0592}{n} \log Q. 2. Calculate Reaction Quotient (Q) Q = \frac Cr^{3+}]^2} Pb^{2+}]^3} = \frac{(0.25)^2}{(0.15)^3}. 3. Determine n (number of electrons transferred) From the balanced equation, n = 6. 4. Apply Nernst Equation E_{cell} = 0.61V - \frac{0.0592}{6} \log \left(\frac{(0.25)^2}{(0.15)^3}\right).

Explanation

1. Identify the Nernst Equation<br /> Use the Nernst equation: $E_{cell} = E_{cell}^{\circ} - \frac{RT}{nF} \ln Q$. For simplicity at room temperature, use $E_{cell} = E_{cell}^{\circ} - \frac{0.0592}{n} \log Q$.<br />2. Calculate Reaction Quotient (Q)<br /> $Q = \frac{[Cr^{3+}]^2}{[Pb^{2+}]^3} = \frac{(0.25)^2}{(0.15)^3}$.<br />3. Determine n (number of electrons transferred)<br /> From the balanced equation, $n = 6$.<br />4. Apply Nernst Equation<br /> $E_{cell} = 0.61V - \frac{0.0592}{6} \log \left(\frac{(0.25)^2}{(0.15)^3}\right)$.
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