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Two half-cell reactions are shown below: Fe2+(aq)  +  2e-   ↔  Fe(s)     Ecell = – 0.44 V Cu2+(aq)  +  2e-  ↔  Cu(s)    Ecell = +0.34 V Hint: The half-cell with the highest positive potential gets reduced. the other half-cell must get oxidized. Find the Gibbs free energy for the above cell reaction (ΔG = -nFE: F = 96485 C, n = no. of electrons and E = cell potential). Use the cell voltage you received in questions 21.

An emission of gamma radiation from a radioactive element 

Consider the following half-cell reactions: Zn2+(aq)  +  2e-  ↔  Zn(s)  Ecell = -0.76 V Cu2+(aq)  +  2e-  ↔  Cu(s)  Ecell = +0.34 V If a cell is built by combining above two half-cells, what will be its cell potential?  Hint: The half-cell with the highest positive potential gets reduced: Ecell= E(cathode) – E(anode)

If Al3+(aq) = 0.36 M, Sn4+ (aq) = 0.086 M and Sn2+(aq) = 0.54 M, calculate the Ecell of this non standard cell Use: Al3+(aq)  +  3e-  ↔  Al(s)            Ecell = -1.676 V Sn4+(aq)  +  2e-  ↔  Sn2+(aq)  Ecell = 0.154 v

Based on the Gibbs free energy you received for question (25), calculate the equilibrium constant for the reaction. Use ΔG = -RT ln K: Temperature 25 Celsius, R = 8.314 J / mol. K

HCl, HBr and HI are strong acids; however, HF is a weak acid. What is the best explanation for this phenomenon?

Embalming practices do not vary from country to country so viewing is rarely a problem when human remains are received from other countries.

What is the concentration of the ions in 3.0 L of 1.0 MAl2(CO3)3?   Answer       [Al 3+]          [CO 3 2 -]      A 0.33 M 0.50 M B 0.66 M 0.99 M C 2.0 M 3.0 M D 3.0 M 4.5 M

Calculate the Ecell for the following reaction? Zn(s)  +  Fe2+(aq, 0.1 M)  ↔  Zn2+(aq, 1.9 M)  +  Fe(s) Use:  Zn2+(aq)  +  2e-  ↔  Zn(s)  Ecell = -0.76 V Fe2+(aq)  = 2e-  ↔  Fe(s)  Ecell = -0.447 V

Predict the products for the following reaction Zn(s)  +  2H+(aq)   →