# What is the cell potential for the reaction Mg(s)+Fe^(2+)(aq)\rarrMg^(2+)(aq)+Fe(s) at 47 °C when [Fe^(2+)]= 3.50 M and [Mg^(2+)]= 0.110 M?

## Notes: To learn how to use the Nernst equation. The standard reduction potentials listed in any reference table are only valid at standard-state conditions of 25 °C and 1 M. To calculate the cell potential at non-standard-state conditions, one uses the Nernst equation, E=E°−(2.303RT)/(nF)\log_10Q where E is the potential in volts, E° is the standard potential in volts, R=8.314J/(K⋅mol) is the gas constant, T is the temperature in kelvins, n is the number of moles of electrons transferred, $F = 96 , 500 \frac{C}{\setminus \textrm{m o l} {e}^{-}}$ is the Faraday constant, and Q is the reaction quotient. Substituting each constant into the equation the result is E=E°-(0.0592 V)/nlog_10Q

May 4, 2017

$1.96 V$

#### Explanation:

Please check if my work is correct!
Extra data:

• Q = $3.14 \setminus \cdot {10}^{- 2}$
• T = 320 K
• n = 2 mol
• E° = 1.92 V

My work:
$E = 1.92 V - \frac{0.0592 V}{2 m o l} \setminus {\log}_{10} \left(3.14 \setminus \cdot {10}^{- 2}\right) \setminus \approx 1.96 V$