Electrochemistry — Flashcards
Punjab Board · Class 12 · Chemistry
36 flashcards for Electrochemistry (Punjab Board Class 12 Chemistry) to test yourself on key terms and facts. Sample: "What is a galvanic cell and how.
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What is a galvanic cell and how does it work?
Answer
A galvanic cell is an electrochemical cell that converts chemical energy from a spontaneous redox reaction into electrical energy. It consists of two half-cells (anode and cathode) connected by a salt…
Differentiate between anode and cathode in a galvanic cell.
Answer
In a galvanic cell: ANODE is the electrode where oxidation occurs, making it the negative electrode (electrons are released here). CATHODE is the electrode where reduction occurs, making it the positi…
Write the half-reactions and overall reaction for the Daniell cell.
Answer
Daniell Cell Half-Reactions: Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻ Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s) Overall Reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) Emf = 1.1 V (at standard cond…
What is the standard hydrogen electrode (SHE) and why is it important?
Answer
The Standard Hydrogen Electrode (SHE) is a reference electrode with a platinum electrode coated with platinum black, dipped in 1 M H⁺ solution, with H₂ gas at 1 bar pressure. It has an assigned potent…
What is meant by standard electrode potential? How do you determine it?
Answer
Standard electrode potential (E°) is the potential of an electrode when all species involved in the half-reaction are at unity concentration (or 1 M) and at 298 K. Determination: Measure the emf of a …
Calculate the emf of a cell formed by Cu and Zn electrodes at standard conditions. (Given: E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = -0.76 V)
Answer
Formula: E°(cell) = E°(cathode) - E°(anode) Step 1: Identify cathode (Cu, higher E°) and anode (Zn, lower E°) Step 2: E°(cell) = 0.34 V - (-0.76 V) Step 3: E°(cell) = 0.34 V + 0.76 V = 1.10 V Answer…
What is the Nernst equation and what does it predict?
Answer
The Nernst equation relates electrode potential to ion concentration at any temperature: E = E° - (RT/nF) ln([reduced]/[oxidized]) Or at 298 K: E = E° - (0.059/n) log([reduced]/[oxidized]) It predi…
Calculate the cell potential for the reaction: Mg(s) + 2Ag⁺(0.0001 M) → Mg²⁺(0.130 M) + 2Ag(s), given E°(cell) = 3.17 V
Answer
Using Nernst Equation: E(cell) = E°(cell) - (0.059/n) log([Mg²⁺]/[Ag⁺]²) Step 1: n = 2 (electrons transferred) Step 2: E(cell) = 3.17 V - (0.059/2) log(0.130/(0.0001)²) Step 3: E(cell) = 3.17 V - 0.0…
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