# 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?

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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=96,500C/(\text(mol )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#

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,

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,

Substituting each constant into the equation the result is

##### 1 Answer

#### Explanation:

**Please check if my work is correct!**

Extra data:

- Q =
#3.14\cdot10^(-2)# - T = 320 K
- n = 2 mol
- E° = 1.92 V

My work: