A certain metal M forms a soluble nitrate salt MNO₃. Suppose the left half cell of a galvanic cell apparatus is filled with a 1. 50 M solution of MNO₃ and the right half cell with a 15. 0 mM solution of the same substance. The right side of the cell is positive and voltage of the cell is 0.12V.
a) The representation of the galvanic cell with a different concentration of MNO₃ is shown as
M(s)|MNO₃(aq)(1.50M)||MNO₃(aq)(15.0mM)|M(s)
On the left hand side of the cell, the oxidation is occurred which is shown as
M(s)→M⁺(aq)+e⁻
Therefore, this electrode is negative as it shows oxidation reaction.
On the right hand side of the cell, the reduction is occurred and it is shown as
M⁺(aq)+e⁻→M(s)
Therefore, this electrode is positive as it shows reduction reaction.
b) The emf of cell is calculated using the formula
E(cell)=E⁰(cell)-RT/nF×ln([reductant/[oxidant])
The concentration of oxidant is 1.50M and of reductant is 15.0mM which is equal to 0.015M and E⁰(cell) is 0(zero) because both are same solution. Hence, the equation will be
E(cell)=-RT/nF×ln([reductant/[oxidant])
Where, T=20⁰C=20+273=293K
n=1
F=96485C
R=8.314J/molK
Plug all values in the formula
E(cell)=-(8.314×293J×mol×K/1mol×96485C×K)×ln(0.015/1.50)
E(cell)=-(2436.002J/96485C)×ln(0.01)
Ecell=-(0.025247×-4.6052)
Ecell=-(-0.1162)J/C
Ecell=0.1162 J/C=0.12V (1V=1J/C)
Therefore, the voltage of the cell is 0.12V.
To know more about galvanic cell here
https://brainly.com/question/23319555
#SPJ4
