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) → Zn^{2+}(aq) + H_{2}(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.

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{-}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}}{\mathbf{n}}\mathbf{\right)}{\mathbf{logQ}}}$

E_{cell} = 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(NO _{3})_{2} dissociates into ions* in solution:

**Zn(NO _{3})_{2} → Zn^{2+} + 2NO_{3}**

This means that adding **Zn(NO _{3})_{2 }**increases [Zn

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) → Zn^{2+}(aq) + H_{2}(g), for each of the following changes?

Zinc nitrate is added to the anode compartment.

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