# Problem: Consider the reaction Mg(s) + Fe2+(aq) → Mg2+(aq)+Fe(s) at 87°C , where [Fe2+]= 3.80M and [Mg2+]= 0.210M.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 forMg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)

###### FREE Expert Solution

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

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

Standard cell potential

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

a) Reaction quotient, Q:

89% (478 ratings) ###### Problem Details

Consider the reaction Mg(s) + Fe2+(aq) → Mg2+(aq)+Fe(s) at 87°C , where [Fe2+]= 3.80M and [Mg2+]= 0.210M.

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) + Fe2+(aq) → Mg2+(aq) + Fe(s)