What is the oxidation half-reaction for mg(s) + zncl2(aq) mgcl2(aq) + zn(s)?

a. zn(s) --> zn^2+ + 2e-
b. mg(s) --> mg2+ + 2e-
c. zn^2+ + 2e- --> zn(s)
d. mg^2+ + 2e- --> mg(s)

B)

Explanation: Oxidation is loss of electrons and reduction is gain of electrons. When an element or ion loses electrons then its positive charge increases and when it gains electrons then charge decreases.

In the given equation, on reactant side Mg has zero charge and Zn has +2 charge. On product side, Mg has +2 charge and Zn has zero charge.

Since the charge of Mg is increasing from zero to +2, its oxidation.

We can ignore choice A and C since none of these have Mg. Choice D could also be ignored as it shows reduction of Mg as its changing from +2 to 0 state.

Hence, the correct choice is B as it shows an oxidation of Mg.

B)

The right answer is Option d (0.00103 M).

Explanation:

The given values are:

T = 31.6

e = 487 L mol cm

length, l = 1 cm

As we know,

Absorbance,

By using the Beer's law, we get

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On substituting the values, we get

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