1. Calculate [Ag + ] for the following electrochemical cell if E cell = 0.67 V P
ID: 881852 • Letter: 1
Question
1. Calculate [Ag+] for the following electrochemical cell if Ecell = 0.67 V
Pt | H2(g), (1.00 atm), H+ (1.00 M) || Ag+ (? M) | Ag
2. How many grams of cadmium are deposited from an aqueous solution of CdSO4when an electrical current of 1.15 A flows through the solution for 312 minutes?
3. Write equations for the following nuclear reactions
a) The radioactive decay of nickel-63 by beta emission.
b) The radioactive decay of polonium-207 by alpha emission.
c) The fission reaction of uranium-235 with neutrons to produce xenon-142, rubidium-95 and positron emission.
4. Calculate the binding energy for the beryllium-9 nucleus given its mass is 9.01218 amu.
5. A 50.0 mg sample of pure scandium-44 is extracted from a large quantity of ore. If the half-life of this isotope is 3.43 days, calculate the amount of scandium-44 remaining 12 days after the extraction.
Explanation / Answer
1. Pt | H2(g), (1.00 atm), H+ (1.00 M) || Ag+ (? M) | Ag
Ag+ + e- ----> Ag ....(1) Eo= 0.80 V
H2 + 2e- ----> 2H+ ...(2) Eo=0.0 V
Multiply eqn 1 by 2 and add both eqns;
2Ag+ + H2 -----> 2Ag+ 2H+
Eo= 0.80-0= 0.80 V
Using Nernst equation; Ecell= Eo-0.0592/n logQ
0.67= 0.80-0.0592/2 log [H+]2/ [Ag+]2 [H2]
0.67-0.80= -0.0296 log(1)/ [Ag+]2(1)
4.39= -log(Ag+)2
log 4.39= -2log[Ag+]
log[Ag+]= -2.196
[Ag+]= 10^-2.196
[Ag+]= 6.37*10^-3 M
2. Current= 1.15 A
Time=312 min= 312*60= 21528 C
Reaction taking place; Cd2+ + 2e- ----> Cd
Thus 2F, i.e., 2*96500 C deposit cadmium = 1 mol= 112.4 gm
21528 C will deposit cadmium= 112.4/ 2*96500 *21528= 12.53 gm
So, 12.53 gm of cadmium is deposited by passing 312 min
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