For the electrochemical cell Zn (s) | Zn2+ (aq) || Cd2+ (aq) | Cd (s), (a) deduce the cell reaction, (b) calculate the standard emf, given that Eo (Zn | Zn2+) = −0.763 V and Eo (Cd | Cd2+) = −0.403 V. [+0.360]

9) At 25°C, what is the potential and the cell reaction for the following cell: Ni | Ni2+ (aq, 0010 M) || Cl(aq, 0.20M) | Cl2 (g, 1 atm) | Pt? What is the Keg at 298K?

Calculate the cell potential at 25°C for the Cr(s)|Cr3+(aq)||Sn²*(aq)|Sn(s) electrochemical cell if the concentration of Cr3+(aq) is o.50 M and the concentration of Sn²+(aq) is o.090 M. Report your answer to 3 significant figures. E cell V

A galvanic cell is constructed as follows. One half-cell consists of a platinum wire immersed in a solution containing 1.00 M Sn2+ and 1.00 M Sn4+. The other half-cell has a thallium rod immersed in a solution of 1.00 M Tl+. What is the equilibrium constant at 25° C?(Eo= Tl+ / Tl = −0.34 V.)? What is the cell voltage if the Tl+ concentration is increased 6−fold?

Mn(s)|Mn2+(aq)||Cr3+(aq)| Cr(s) Calculate the cell potential at 25oC if the concentration of Mn2+(aq) is 2.40 M and the concentration of Cr3+(aq) is 0.500 M. Report your answer to 3 significant figures. Use theStandard Reduction Potentials Table as needed.

3. At 298 K, the EMF of the cell shown below is +0.84 V. Pt(s)|Fe²*(aq, a = 1), Fe³*(aq, 1)|| Ce**(aq, a = 1), Ce³*(aq, a = 1)|Pt(s) a = (i) Define what is meant by the standard EMF of the cell. (ii) Write down the cell reaction and the Nernst equation for the cell. (iii) Calculate the equilibrium constant for the cell reaction at 298 K.

For the following electrochemical cell: Mn(s)|Mn2+(aq)(.215M)||Cr3+(aq)(.113M), Cr2+(aq)(.301M)|Pt(s) Determine the cell potential for this electrochemical cell by first calculating the standard cell potential and then by using the Nernst equation.

Which of the following reactions occurs at the anode of an electrochemical cell composed of Zn│Zn2+(aq) and Fe│Fe2+(aq) half-cells? Their reduction potentials are as follows:                 Zn2+   +  2e–   -->  Zn;    Eo = -0.763 V                              Fe2+  +  2e–   -->  Fe;      Eo = -0.447 V     (A) Zn2+ + 2e– -->  Cr;                     (B) Zn -->  Zn2+ + 2e–;                    (C) Fe -->  Fe2+ + 2e–;              (D) Fe2+ + 2e– -->  Fe;