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1876 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

1876 United States presidential election in Rhode Island

← 1872 November 7, 1876 1880 →
Nominee Rutherford B. Hayes Samuel J. Tilden
Party Republican Democratic
Home state Ohio New York
Running mate William A. Wheeler Thomas A. Hendricks
Electoral vote 4 0
Popular vote 15,787 10,712
Percentage 59.29% 40.23%

President before election

Ulysses S. Grant

Elected President

Rutherford B. Hayes

The 1876 United States presidential election in Rhode Island took place on November 7, 1876, as part of the 1876 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island voted for the Republican nominee, Rutherford B. Hayes, over the Democratic nominee, Samuel J. Tilden. Hayes won the state by a margin of 19.06%.

With 59.29% of the popular vote, Rhode Island would be Hayes' fourth strongest victory in terms of percentage in the popular vote after Vermont, Nebraska and Kansas.[1]

This was the only election between 1860 and 1888 where the Democratic candidate earned more than 40% in at least one Rhode Island county.


1876 United States presidential election in Rhode Island[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Rutherford B. Hayes of Ohio William A. Wheeler of New York 15,787 59.29% 4 100.00%
Democratic Samuel J. Tilden of New York Thomas A. Hendricks of Indiana 10,712 40.23% 0 0.00%
Greenback Peter Cooper of New York Samuel Fenton Cary of Ohio 68 0.26% 0 0.00%
Prohibition Green Clay Smith of Kentucky Gideon Tabor Stewart of Ohio 60 0.23% 0 0.00%
Total 26,627 100.00% 4 100.00%


  1. ^ "1876 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1876 Presidential General Election Results - Rhode Island".

This page was last edited on 11 July 2019, at 18:50
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