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kipiarov [429]
3 years ago
5

Consider the following half reactions and their standard reduction potential values to answer the following questions.

Chemistry
1 answer:
Gennadij [26K]3 years ago
8 0

Answer:

Cu2+ (aq) + e- -------->Cu+ (aq)​E° = 0.15 V anode

Br2 (l) + 2e- -------------> 2Br- (aq) ​E° = 1.08 V cathode

Explanation:

The half equation having a more positive reduction potential indicates the reduction half equation while the half equation having the less positive reduction potential indicates the oxidation half equation.

The overall redox reaction equation is;

2Cu^+(aq) + Br2(g) ----> 2Cu^2+(aq) + 2Br^-(aq)

E°cell= E°cathode -E°anode

E°cathode= 1.08 V

E°anode= 0.15 V

E°cell= 1.08V-0.15V

E°cell= 0.93V

From;

∆G=-nFE°cell

n= 2

F=96500C

E°cell= 0.93V

∆G= -(2×96500×0.93)

∆G= - 179.49 J

From ∆G= -RTlnK

∆G= - 179.49 J

R=8.314Jmol-K-1

T= 25° +273= 298K

K= ???

lnK= ∆G/-RT

lnK=-( - 179.49/(8.314×298))

lnK= 0.0724

K= e^0.0724

Kc= 1.075

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How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?
zlopas [31]
<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.60 × 10⁻⁵ g Au (Gold)

<u>Step 2: Identify Conversions</u>

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
  2. Multiply:                            \displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

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