Question

A galvanic cell is set up with two cells containing a cadmium electrode in 0. 250 m cd(no3)2 solution and an iron electrode in 0. 050 m fe(no3)2 solution. what is the cell potential at 25 oc?

Answer 1

Cd2+(aq) + 2e- ---> Cd(s) , E° = -0.40 V Fe2+(aq) + 2e- ----> Fe(s) , E° = -0.44 V Since standard reduction potential.

Galvanic Cell: A chemical cell, also known as an electrochemical cell, is a device that transforms the chemical energy generated during a red-ox reaction into electrical energy. In honor of Luigi Galvanic and Alessandro Volta, who conducted the first experiments on the conversion of chemical energy into electrical energy, these are also known as galvanic or voltaic cells.

A spontaneous red-ox reaction occurs in an electrochemical cell in an indirect manner, and the reduction in free energy that occurs during the chemical reaction manifests as electrical energy. In an indirect red-ox reaction, the reduction and oxidation processes take place in two different containers known as half-cells.

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Answer 2

The cell potential is 0.01934 V at 25°C.

What is a galvanic cell?

A galvanic cell, also known as a voltaic cell, is an electrochemical cell that converts the chemical energy generated by spontaneous redox reactions into electrical energy.

The main components of a galvanic cell are

Anode – Oxidation occurs at the anode

Cathode – Reduction occurs at the cathode

Salt bridge – consists of electrolytes that are needed to complete the circuit in a galvanic cell.

Half-cells – It is important as reduction and oxidation reactions are separated into compartments.

External circuit – it helps in Conducting the flow of electrons between electrodes

At cathode,

$Cd^{2+}(aq) + 2e^-\rightarrow Cd(s)$

At anode,

$Fe(s)+2e^-\rightarrow Fe^{2+}(aq)$

Overall cell reaction

$Fe(s)+Cd^{2+}(aq)\rightarrow Fe^{2+}(aq) + Cd(s)$

$E^o_{cell} = E^o_{cathode} - E^o_{anode}$

          = -0.40 -(0.44) = 0.04 V

Nernst equation is given by

$E_{cell} = E^o_{cell} - \frac{2.303RT}{nF} logQ$

At 25°C

$E_{cell} = E^o_{cell} - \frac{0.0592V}{n} logQ$

Here, n = 2 and log Q = log $\frac{0.250}{0.050}$ = 0.699

$E_{cell}$  = 0.01934 V

The cell potential is 0.01934 V at 25°C

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