1880 United States presidential election in New Jersey
The 1880 United States presidential election in New Jersey took place on November 2, 1880, as part of the 1880 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for president and vice president.
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Elections in New Jersey |
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New Jersey voted for the Democratic nominee, Winfield Scott Hancock, over the Republican nominee, James A. Garfield. Hancock won the state by a very narrow margin of 0.82%.
A Democratic candidate would not win New Jersey while losing the national popular vote again until 2004.
Results
1880 United States presidential election in New Jersey[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Democratic | Winfield Scott Hancock of Pennsylvania | William Hayden English of Indiana | 122,565 | 49.84% | 9 | 100.00% | ||
Republican | James Abram Garfield of Ohio | Chester Alan Arthur of New York | 120,555 | 49.02% | 0 | 0.00% | ||
Greenback | James Baird Weaver of Iowa | Barzillai Jefferson Chambers of Texas | 2,617 | 1.06% | 0 | 0.00% | ||
Prohibition | Neal Dow of Maine | Henry Adams Thompson of Ohio | 191 | 0.08% | 0 | 0.00% | ||
Total | 245,928 | 100.00% | 9 | 100.00% |
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