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Elections in Vermont |
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The 1876 United States presidential election in Vermont took place on November 7, 1876, as part of the 1876 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.
Vermont voted for the Republican nominee, Rutherford B. Hayes, over the Democratic nominee, Samuel J. Tilden. Hayes won Vermont by a margin of 36.92%.
With 68.30% of the popular vote, Vermont would be Hayes' strongest victory in terms of percentage in the popular vote. [1]
1876 United States presidential election in Vermont [2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Rutherford B. Hayes of Ohio | William A. Wheeler of New York | 44,091 | 68.30% | 5 | 100.00% | ||
Democratic | Samuel J. Tilden of New York | Thomas A. Hendricks of Indiana | 20,254 | 31.38% | 0 | 0.00% | ||
N/A | Others | Others | 208 | 0.32% | 0 | 0.00% | ||
Total | 64,553 | 100.00% | 5 | 100.00% |
The 1876 United States presidential election was the 23rd quadrennial presidential election, held on Tuesday, November 7, 1876, in which Republican nominee Rutherford B. Hayes faced Democrat Samuel J. Tilden. It was one of the most contentious presidential elections in American history. Its resolution involved negotiations between the Republicans and Democrats, resulting in the Compromise of 1877, and on March 2, 1877, the counting of electoral votes by the House and Senate occurred, confirming Hayes as President. It was the second of five U.S. presidential elections in which the winner did not win a plurality of the national popular vote.
William Almon Wheeler was an American politician and attorney. He served as a United States representative from New York from 1861 to 1863 and 1869 to 1877, and the 19th vice president of the United States from 1877 to 1881.
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