A) Based on the following information, Br 2(l) + 2 e - 2 Br - ( aq ) E ° = +1.09

Question

A) Based on the following information,
Br 2(l) + 2 e- 2 Br -(aq) E° = +1.09 V
Mg 2+(aq) + 2 e- 2 Mg(s) E° = -2.37 V
which of the following chemical species is the strongest reducing agent?

A) 2.23 B) 2.2 C) 2.32 D) 2.39 E) 2.1

3) In an electrochemical cell, Q=25 and K=0.300. What can you conclude about Ecell and Ecell?
A) Ecell and Ecell are both positive.
B) Ecell is negative and Ecell is positive.
C) Ecell and Ecell are both negative.
D) Ecell is positive and Ecell is negative.

4) Ni2+(aq)+2eNi(s)E=0.23V
Zn2+(aq)+2eZn(s)E=0.76V
Use appropriate data to determine which statement is true of the voltaic cell picture here.
A) Zn is the anode; Ni is the cathode; electrons flow from right to left
B) Zn is the cathode; Ni is the anode; electrons flow from right to left
C) Zn is the anode; Ni is the cathode; electrons flow from left to right
D) Zn is the cathode; Ni is the anode; electrons flow from left to right

a) Br -(aq)
b) Br 2(/)
c) Mg 2 + (aq)
d) Mg (s)
2) The standard emf for the cell using the overall cell reaction below is +2.20 V:
2Al(s) + 3I2(s) 2Al3+(aq) + 6I-(aq)
The emf generated by the cell when [Al3+] = 4.5 × 10-3 M and [I-] = 0.15 M is

A) 2.23 B) 2.2 C) 2.32 D) 2.39 E) 2.1

Explanation / Answer

1) we know that

lower the reduction potential stronger the reducing agent

so

looking at the values

we can see that

Mg (s) is the best reducing agent

option d) Mg (s)

2)

we know that

according to nernst equation

E = Eo - ( 0.0591/n) log Q

now

the reaction is

2 Al (s) + 3 I2 (s) ---> 2 Al+3 (aq) + 6I- (aq)

reaction quotient is given by

Q = [Al+3]^2 [ I-]^6

Q = [ 4.5 x 10-3 ]^2 [0.15]^6

Q = 2.3 x 10-10

now

E = Eo - ( 0.0591/n) log Q

E = 2.2 - ( 0.0591/6 ) log 2.3 x 10-10

E = 2.32

so

the answer is option C) 2.32

3)

we know that

Eo= (RT/nF) lnK

Eo (RT/nF) ln 0.3

Eo = -1.2 (RT/nF)

so

Eo cell is negative

now

E= Eo - ( 0.0591/n) log Q

given

Q = +ve

also Eo = -ve

so

E = -ve

so

the answer is

C) E and Eo are both negative

4)

we know that

lower potential is anode

so

Zn is anode and Ni is cathode

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