Consider this reaction. 2Fe^3+ (aq) + 2I degree (aq) rightarrow 2Fe^2+ (aq) + I_
ID: 1071425 • Letter: C
Question
Consider this reaction. 2Fe^3+ (aq) + 2I degree (aq) rightarrow 2Fe^2+ (aq) + I_2 (aq) Which statement is true fat the reaction? Fe^3+ is oxidized. Fe^3+ increases in oxidation number Fe^3+ is reduced. I^- is reduced. Which statement about this redox reaction is correct? 2MnO_4^- (aq) + 5H_2 O_2 (aq) + 6H^+ (aq) rightarrow 2Mn^2+ (aq) + 5O_2 (g) + 8H_2 O(g) O_2 acts as the oxidant in this reaction. H_2 O_2 acts as an oxidizing agent. H_2 O_2 acts as a reducing agent. Only oxidation has taken place. An oxidation-reduction reaction in which 3 electrons are transferred has Delta G degree = +18.55 kJ at 25 degree C. What is the value of E degree? +0.192 V -0.064 V -0.192 V -0.577 V What time is required to plate 2.08 g of copper at a constant current flow of 1.26 A? Cu^2+ (aq) + 2e rightarrow Cu(x) 41.8 min 83.6 min 133 min 5016 min Consider this reaction. Cu^2+ (aq) + Fe(s) rightarrow Cu(s) + Fe^2+ aq E degree = 0.78 V What is the value of E when [Cu^2+]Explanation / Answer
1. Fe3+ is reduced. The answer is C
2. H2O2 acts a reducing agent. The answer is B
3. G = -nFE
G = 18.55 kJ = 18550 J
n = 3
E = -G/nF = -18550 J/ 3x96485 C mol-1 = -0.064 V
The answer is B
4. Cu = 2.08 g
Electrochemical equivalent = 63.5/2 = 31.75g
No of equivalent in 2.08 g = 2.08/31.75 = 0.0655
A = 1.26 A
The time required to deposit the amountof Cu in t seconds
no. of Faradays = no. of equivalents
(1.26xt/96485) = 0.0655
t = 0.0655 x 96485/1.26 = 5015.6 sec
Time in minutes = 5015.6/60 = 83.6 min
The answer is B
5.
E = Eo- (0.0592/n) logQ
Eo = 0.78
n = 2
log Q = [Fe2+]/[Cu2+] = 0.40/0.040 = 10
E = Eo- (0.0592/n) logQ
E =0.78 - (0.0592/2) log10
E = 0.78 - 0.0296x1 = 0.75 V
The answer B