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=96,500C/(\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:
