Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃] = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
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Answer: a) Anode: 
Cathode: 
b) Anode : Cr
Cathode : Au
c) 
d) 
Explanation: -
a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
At cathode: 
At anode: 
b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.
At anode which is a negative terminal, oxidation occurs which is loss of electrons.
Gold acts as cathode ad Chromium acts as anode.
c) Overall balanced equation:
At cathode:
(1)
At anode:
(2)
Adding (1) and (2)

d)
= -0.74 V
= 1.40 V

Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAu%5E%7B3%2B%7D%5D%7D%7B%5BCr%5E%7B3%2B%7D%5D%5E%7D)
where,
n = number of electrons in oxidation-reduction reaction = 3
= standard electrode potential = 2.14 V
![E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.14-%5Cfrac%7B0.0592%7D%7B3%7D%5Clog%20%5Cfrac%7B%5B1.0%7D%7B%5B1.0%5D%7D)

Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.
There are 20 milligrams in 0.2 decigrams