Good idea!maybe I should try that
Zinc is no longer the positive electrode because copper has a more positive (higher) value than zinc (anode). The anode value is reduced by the potential of the other electrode.
<h3>In a galvanic cell, is the anode positive or negative?</h3>
In a galvanic (voltaic) cell, the cathode is regarded as positive and the anode as negative. This seems reasonable given that the cathode is where electrons flow from the anode, which is where they originate.
<h3>What is a galvanic cell?</h3>
An electrochemical cell called a galvanic cell or voltaic cell, respectively named after the scientists Luigi Galvani and Alessandro Volta produces an electric current by spontaneous oxidation-reduction reactions. A typical device typically consists of two distinct metals that are submerged in separate beakers that each contains their own metal ions in solution and are either connected by a salt bridge or divided by a porous membrane.
Learn more about Galvanic cells here:-
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Answer:
A. the burning of fossil fuels within the Sun
Explanation:
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻
1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions
the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol
NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol
total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³
Answer:
Explanation:
There are two ways of looking at this problem. The first way, slightly more advanced, is to understand that the carbocation formed is an intermediate in this reaction: it is formed in one step and consumed in the subsequent step.
Secondly, we have hydroxide involved as our reactant, so it should be our second reactant in the second bimolecular step.
Thirdly, the product formed would be a combination of the anion and cation, one of our products, this means we have the following second step:
Another way is to verify this knowing that by adding all of the steps should yield a net equation, notice if we add the two steps together (reactants on one side and products on the other), we obtain:
Notice that the intermediate carbocation cancels out on both sides to yield the final net equation:
This means we have the correct second step.
