The **reaction quotient, Q**, can be calculated like an equilibrium constant with the formula shown below:

$\overline{){\mathbf{Q}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}$

Conversion from** °C to K**:

$\overline{){\mathbf{K}}{\mathbf{=}}{\mathbf{\xb0}}{\mathbf{C}}{\mathbf{+}}{\mathbf{273}}}$

**Standard cell potential**:

$\overline{){\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cathode}}}{\mathbf{-}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{anode}}}}$

**a) R****eaction quotient, Q:**

$\mathbf{Q}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}\mathbf{=}\frac{\left[{\mathrm{Mg}}^{2+}\right]}{\left[{\mathrm{Fe}}^{2+}\right]}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{Q}\mathbf{=}\frac{[0.210\overline{)M}]}{[3.80\overline{)M}]}$

Consider the reaction Mg(s) + Fe^{2+}(aq) → Mg^{2+}(aq)+Fe(s) at 87°C , where [Fe^{2+}]= 3.80*M* and [Mg^{2+}]= 0.210*M.*

What is the value for the reaction quotient, *Q*, for the cell?

What is the value for the temperature, *T*, in kelvins?

What is the value for *n*?

Calculate the standard cell potential for

Mg(s) + Fe^{2+}(aq) → Mg^{2+}(aq) + Fe(s)

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Based on our data, we think this problem is relevant for Professor Edinger's class at UCI.