John M. answered • 12/11/17

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H.S. Chemistry, A.P. Chem, Physical Science and College Freshman Chem

The equation is: E

_{cell}= E^{o }_{cell }- (0.0592/n) log QYou know all values except Q (n = 2). Solve for Q. This is the ratio of the concentration of the products over the reactants, raised to the power of their coefficients. Since, Ag and Zn are solids, their concentrations are constant and do not enter into the calculation.

The only conundrum is whether the Zn2+ is a product or reactant. Based on the given info, it says that a Zn wire is attached to the positive terminal: the anode. Oxidation occurs at the anode, so the oxidation reaction is Zn --> Zn

^{2+}+ 2 e^{-}. So, the Q ratio is [Zn^{2+}]/ unity. Hence, 10^{Q}= [Zn2^{]}