United States presidential election in Rhode Island, 1828

United States presidential election in Rhode Island, 1828

October 31 – December 2, 1828

Nominee John Quincy Adams Andrew Jackson Party National Republican Democratic Home state Massachusetts Tennessee Running mate Richard Rush John C. Calhoun Electoral vote 4 0 Popular vote 2,754 821 Percentage 77.03% 22.97%

Elections in Rhode Island
Federal government Presidential Elections 1792 1796 1800 1804 1808 1812 1816 1820 1824 1828 1832 1836 1840 1844 1848 1852 1856 1860 1864 1868 1872 1876 1880 1884 1888 1892 1896 1900 1904 1908 1912 1916 1920 1924 1928 1932 1936 1940 1944 1948 1952 1956 1960 1964 1968 1972 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 Presidential Primaries Democratic 2008 Republican 2008 2012 U.S. Senate elections 1970 1972 1976 1978 1982 1984 1988 1990 1994 1996 2000 2002 2006 2008 2012 2014 U.S. House elections 2006 2008 2010 2012 2014 2016
State government Gubernatorial Elections 1994 1998 2002 2006 2010 2014

The 1828 United States presidential election in Rhode Island took place between October 31 and December 2, 1828, as part of the 1828 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 National Republican candidate, John Quincy Adams, over the Democratic candidate, Andrew Jackson. Adams won Rhode Island by a margin of 54.06%.

Results

References

State Results of the 1828 U.S. presidential election
Candidates	Andrew Jackson John Quincy Adams
General articles	Election timeline
Local results	Alabama Connecticut Delaware Georgia Illinois Indiana Kentucky Louisiana Maine Maryland Massachusetts Mississippi Missouri New Hampshire New Jersey New York North Carolina Ohio Pennsylvania Rhode Island South Carolina Tennessee Vermont Virginia
Other 1828 elections	House Senate Gubernatorial