Question

A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction: →+3Cu+2aq2Als+3Cus2Al+3aq Suppose the cell is prepared with 7.97 M Cu+2 in one half-cell and 2.50 M Al+3 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

Answer 1

Answer:

Cell voltage under these condition is 2.02 V.

Explanation:

(Oxidation: $Al-3e^{-}\rightarrow Al^{3+}$)$\times 2$ ; $E_{Al^{3+}\mid Al}^{0}=-1.66V$

(Reduction: $Cu^{2+}+2e^{-}\rightarrow Cu$)$\times 3$ ; $E_{Cu^{2+}\mid Cu}^{0}=0.34V$

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Overall: $2Al+3Cu^{2+}\rightarrow 2Al^{3+}+3Cu$

Nernst equation for this cell reaction at $25^{0}\textrm{C}$-

$E_{cell}=[E_{Cu^{2+}\mid Cu}^{0}-E_{Al^{3+}\mid Al}^{0}]-\frac{0.059}{n}log\frac{[Al^{3+}]^{2}}{[Cu^{2+}]^{3}}$

Where, n is number of electron exchanged and species under third bracket represents concentration. Concentration of solids and liquids are excluded.

So, plug-in all the values in the above equation-

$E_{cell}=(0.34+1.66)-\frac{0.059}{6}log\frac{(2.50)^{2}}{(7.97)^{3}}V$

or, $E_{cell}=2.02V$

So, cell voltage under these condition is 2.02 V