Redox reactions are basically oxidation-reduction reactions that involves in transfer of electrons from one reactant to another. When an atom loses an electron, that process is known as oxidation and the substance that loses electrons is called as reducing agents. When an atom gains an electron, that process is reduced and the substance that gain electrons are called oxidizing agents.

For example: $2Fe^{3+} + Sn^{2+} -> 2Fe^2+ + Sn^{4+}$

Here you can see that $Fe^{3+}$ is reduced to $Fe^{2+}$ and $Sn^{2+}$ is oxidised to $Sn^{4+}$.

This online Redox reaction calculator will calculate the redox reaction potential by the formula given below,

$E=E^{\circ}

$ - $\frac{RT}{nF}$$lnQ$

where,

$E^{\circ}$ = Standard potential

R = Gas constant

F = Faraday constant

n = number of electrons

Q = $\frac{C^c \ D^d}{A^a \ B^b}$

They are coefficients for the reaction :

aA + bB --> cC + dD

here,

uppercase letters = concentration

lowercase letters = stoichiometric

For example: $2Fe^{3+} + Sn^{2+} -> 2Fe^2+ + Sn^{4+}$

Here you can see that $Fe^{3+}$ is reduced to $Fe^{2+}$ and $Sn^{2+}$ is oxidised to $Sn^{4+}$.

This online Redox reaction calculator will calculate the redox reaction potential by the formula given below,

$E=E^{\circ}

$ - $\frac{RT}{nF}$$lnQ$

where,

$E^{\circ}$ = Standard potential

R = Gas constant

F = Faraday constant

n = number of electrons

Q = $\frac{C^c \ D^d}{A^a \ B^b}$

They are coefficients for the reaction :

aA + bB --> cC + dD

here,

uppercase letters = concentration

lowercase letters = stoichiometric

Step 2: Using the redox reaction formula given above, substitute the given values to get redox reaction.

Below are some of the problems based on redox reaction.

### Solved Examples

**Question 1:**For formation of potassium dichromate solution, the $E^{\circ}$ value is -0.03 V and the number of electrons transferred is 2 at 290K. Determine the value of E, if the Q value is 32.46.

**Solution:**

Step 1: Given:

$E^{\circ}$ = -0.03 V

n = 2

Q = 32.46

T = 290K

Step 2:

$E=E^{\circ}$ - $\frac{RT}{nF}$lnQ$

E= -0.03 V -$\frac{8.314 J/mol K \times 290 k}{2 \times 96485.309 C/mol}$$ln 32.46$

E=-0.073 V

Therefore, -0.073 is the redox rection potential.

**Question 2:**For formation of acetaldehyde the $E^{\circ}$ value is -0.580 V and the number of electrons transferred is 2 at 270k. Determine the value of E, if the Q value is 20.15.

**Solution:**

Step 1: Given:

$E^{\circ}$ = -0.580 V

n = 2

Q = 20.15

T = 270K

Step 2:

$E=E^{\circ}$ -$\frac{RT}{nF}$$lnQ$

$E= -0.580 V - $\frac{8.314 J/mol K \times 270 k}{2 \times 96485.309 C/mol}$$ln 20.15$

E= -0.615 V

Therefore, -0.615 is the redox rection potential.