# Problem: What is the effect on the emf of the cell shown in Figure 20.9 in the textbook, which has the overall reactionZn(s) + 2H+(aq)  →  Zn2+(aq) + H2(g), for each of the following changes?Zinc nitrate is added to the anode compartment.

###### FREE Expert Solution

We’re asked to determine the effect of adding Zinc nitrate to the anode compartment on the emf of the cell which has the overall reaction:

Zn(s) + 2H+(aq)    Zn2+(aq) + H2(g)

We will use the Nernst Equation to determine the change in the cell potential with the given conditions. The Nernst Equation relates the concentrations of compounds and cell potential.

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{°}}}_{{\mathbf{cell}}}{\mathbf{-}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}}{\mathbf{n}}\mathbf{\right)}{\mathbf{logQ}}}$

Ecell = cell potential under non-standard conditions
E°
cell = standard cell potential
n = number of e
- transferred
Q= reaction quotient = [products]/[reactants]

The reaction quotient, Q for the overall reaction is:

$\overline{){\mathbf{Q}}{\mathbf{=}}\frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]}}$

$\mathbf{Q}\mathbf{=}\frac{\left[{\mathrm{Zn}}^{2+}\right]}{{\left[{H}^{+}\right]}^{2}}$

*recall that Q is like an equilibrium expression  solids and liquids are not included
*the coefficients of the reactants/products will be their exponents in the Q expression

Zinc Nitrate, Zn(NO3)2 dissociates into ions in solution:

Zn(NO3)2 → Zn2+ + 2NO3-

This means that adding Zn(NO3)2  increases [Zn2+] concentration

81% (415 ratings) ###### Problem Details

What is the effect on the emf of the cell shown in Figure 20.9 in the textbook, which has the overall reaction
Zn(s) + 2H+(aq)  →  Zn2+(aq) + H2(g), for each of the following changes?

Zinc nitrate is added to the anode compartment.